Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 6 Gaseous State Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

11th Chemistry Guide Gaseous State Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the intermolecular interactions become significant
Answer:
(d) at high pressure the intermolecular interactions become significant

Question 2.
Rate of diffusion of a gas is
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) [P + \(\frac{a}{n^{2} V^{2}}\)](V – nb) = nRT

(b) [P + \(\frac{n a}{n^{2} V^{2}}\)](V – nb) = nRT

(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

(d) [P + \(\frac{n^{2} a^{2}}{V^{2}}\)(V – nb) = nRT]
Answer:
(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) collide without loss of energy
Answer:
(b) exert no attractive forces on each other

Question 5.
Equal weights of methane and oxygen are mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{3}\) × 273 × 298
Answer:
(a) \(\frac{1}{3}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature

Question 7.
In a closed room of 1000 m3 a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle

Question 9.
The value of the universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of pressure and volume
Answer:
(d) units of pressure and volume

Question 10.
The value of the gas constant R is
(a) 0.082 dm3atm
(b) 0.987 cal mol-1K-1
(c) 8.3J mol-1K-1
(d) 8erg mol-1K-1
Answer:
(c) 8.3J mol-1K-1

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 11.
The use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Question 12.
The table indicates the value of vanderWaals constant ‘a’ in (dm3)2 atm. mol-2. The gas which can be most easily liquefied is

Gas O2 N2 NH3 CH4
A 1.360 1.390 4.170 2.253

(a) O2
(b) N2
(c) NH3
(d) CH4
Answer:
(c) NH3

Question 13.
Consider the following statements
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises
Select the correct statement
(a) I and II
(b) II and III
(c) I and III
(d) I, II, and III
Answer:
(b) II and III

Question 14.
The compressibility factor for CO2 at 400 K and 71.0 bar is 0.8697. The molar volume of CO2 under these conditions is
(a) 22.04 dm3
(b) 2.24 dm3
(c) 0.41 dm3
(d) 19.5 dm3
Answer:
(c) 0.41 dm3

Question 15.
If the temperature and volume of an ideal gas is increased to twice its values the initial pressure P becomes
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(b) 2P

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 times that of a hydrocarbon having molecular formula CnH2n – 2. What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(d) 1

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape.
(a) \(\frac{3}{8}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{4}\)
Answer:
(c) \(\frac{1}{8}\)

Question 18.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion i.e., α = 1\(\left[\frac{\partial V}{\delta T}\right]\)Vp. For an ideal gas, α is equal to
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(a) T

Question 19.
Four gases P, Q, R, and S have almost the same values of ‘b’ but their a’ values (a. h are Vander Waals Constants) are in the order Q < R < S < p. At a particular temperature, among the four gases, the most easily liquelìable one is
(a) P
(b) Q
(c) R
(d) S
Answer:
(c) R

Question 20.
Maximum deviation from ideal gas is expected
(a) CH4(g)
(b) NH3(g)
(c) H2 (g)
(d) N2 (g)
Answer:
(b) NH3(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 21.
The units of Vander Waals constants ‘b’ and ‘a’ respectively
(a) mol L-1 and L atm2 mol-1
(b) mol L and L atm mol2
(c) mol-1 L and L2 atm mol-2
(d) none of these
Answer:
(c) mol-1 L and L2 atm mol-2

Question 22.
Assertion:
Critical temperature of CO2 is 304 K. it can be liquefied above 304 K.
Reason:
For a given mass of gas, volume is to directly proportional to pressure at constant temperature
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 23.
What is the density of N2 gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K-1 mol-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured in K)
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 1
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 2

Question 25.
25g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

II. Answer these questions briefly:

Question 26.
State Boyle’s law.
Answer:
Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law.
Answer:
Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. The volume of balloon decreases when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.

Question 28.
Name two items that can serve as a model for ‘Gay Lusaac’ law and explain.
Answer:
Firing a bullet:
When gunpowder burns, it creates; a significant amount of superheated gas. The high pressure of the hot gas behind the bullet forces it out, of the barrel of the gun.
Heating food in an oven:
When you keep food in an oven for heating, the air inside the oven is heated, thus pressurized.

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what j the mathematical expression means.
Answer:

  1. The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
  2. \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
    Where V1 and n1 are the volume and number of moles of a gas and V2 and n2 are the values of volume and number of moles of same gas at a different set of conditions.
  3. If the volume of the gas increase then the number of moles of the gas also increases.
  4. At a certain temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 30.
What are ideal gases? In what way real gases differ from ideal gases?
Answer:
The kinetic theory of gases which is the basis for the gas equation (PV = nRT), assumes that the individual gas molecules occupy negligible volume when compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases.

But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases f tend to approach the ideal behaviour under certain conditions.

Question 31.
Can a Vander Waals gas with a = 0 be liquefied? Explain.
Answer:

  • a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
  • If a = 0, there is a very less interaction between the molecules of gas.
  • ‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
  • If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Question 32.
Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:
Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept underwater during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The tyre of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up to a larger altitude.
Answer:
(a) In aerated water bottles, CO2 gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more gas will be present above the liquid surface in the glass bottle.

In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept underwater. As a result, the temperature is likely to decrease and the solubility of CO2 is likely to increase in aqueous solution resulting in decreased pressure.

(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes an accident.

However, if the bottle is cooled under tap water for some time, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will come out of the bottle at a slower rate, reduces the chances of an accident.

(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.

(d) The volume of the gas is inversely proportional to the pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Question 34.
Give a suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container.
Answer:
According to kinetic theory, gas molecules are moving continuously at random. Hence, they do not settle at the bottom of a container.

(b) Gases diffuse through all the space available to them.
Answer:
Gases have a tendency to occupy all the available space. The gas molecules migrate from a region of higher concentration to a region of lower concentration. Hence, gases diffuse through all the space available to them.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 35.
Suggest why there is no hydrogen in our atmosphere. Why does the moon have no atmosphere?
Answer:
1. Hydrogen is the lightest element thus when produced in a free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There is literally leaks from the atmosphere to the empty space. Hydrogen easily gains the velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O2, in its way to produce H2O. So majority portion of H2 reacts and a very less amount of it present in the upper level of the atmosphere and gains velocity to escape the atmosphere.

2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon has thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere on the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that the moon has no atmosphere.

Question 36.
Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if
(a) it is compressed to a smaller volume at f constant pressure.
Answer:
When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior

(b) the temperature is raised while keeping the volume constant.
Answer:
When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally.

(c) More gas is introduced into the same volume and at the same temperature.
Answer:
When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.

Question 37.
Which of the following gases would you expect to deviate from ideal behavior under conditions of low-temperature F2, Cl2, or Br2? Explain.
Answer:
1. Bromine deviates (Br2) from the ideal gas maximum than Cl2 and F2. Because Br2 has the biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.

2. Br2 deviates from ideal behaviour because it has the largest atomic radii compared to Cl2 and F2. So it contains more electrons than the other two, and the Vander Waals forces are stronger in Br2 than in Cl2 and F2. So Br2 deviates from ideal behaviour.

Question 38.
Distinguish between diffusion and effusion.
Answer:

Diffusion Effusion
1. The property of gas which involves the movement of the gas molecules through another gas is called diffusion. It is the property in which a gas escapes from a container through a very small hole.
2. It is the ability of gases to mix with each other It is the ability of gas to travel through a small hole.
3. The rate of diffusion of a gas depends on how fast the gas molecules are moving The rate that this happens depends on how many gas molecules “collide” with the pore.
4. e.g., Smell of perfume diffuses into the air e.g., Air escaping slowly through the pinhole in a tire.

Question 39.
Aerosol cans carry a clear warning of heating of the can. Why?
Answer:
If aerosol cans are heated, then they will produce more vapour inside the can, which will make the pressure rise very quickly. The rise in temperature can double the pressure inside. Even though the cans are tested, they will burst if the pressure goes up too far. A bursting can could be dangerous.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 40.
When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward the back of the car. Upon forward acceleration, the passengers are pushed toward the front of the car. Why?
Answer:
1. When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car due to the inertia of the body, but a helium balloon pushed toward the back of the car. A helium balloon responds to the air around it. Helium molecules are lighter than the air of our atmosphere, and so they move toward the back by gravity as a result of the accelerating frame.

2. Upon forwarding acceleration, the passenger’s arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. A helium balloon is going to move opposite to this pseudo gravitational force.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and Volume.
Answer:
Vander Waals equation for a real gas is given by
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Where n is the number of moles of gas and V is the volume of the container
⇒ P ∝ \(\frac{n^{2}}{V^{2}}\)
⇒ P = \(\frac{a n^{2}}{V^{2}}\)
Where a is proportionality constant and depends on the nature of gas
Therefore, P = P + \(\frac{a n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\) π(2r)3 = 8Vm
where Vm is a volume of a single molecule.
Excluded volume for single molecule = \(\frac{8 V_{m}}{2}\) = 4Vm
Excluded volume for n molecule
= n(4Vm) = nb
Where b is van der Waals constant which is equal to 4Vm
⇒ V’ = nb
Videal = V – nb
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 43.
Derive the values of van der Waals equation constants in terms of critical constants.
Answer:
The Van der walls equation for n moles is
(p + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT ……… (1)
For 1 mole
(P + \(\frac{n^{2}}{V^{2}}\)) (V – b) = RT ……………(2)

From the equation we can derive the values of critical constants Pc, Vc, and Tc in terms of a and b, the van der Waals constants, On expanding the above equation,
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0 ………….(3)

Multiply equation (3) by \(\frac{V^{2}}{P}\)
\(\frac{V^{2}}{P}\) (PV +\(\frac{a}{V}\) – pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V3 + \(\frac{a}{P}\)V – bV2 – \(\frac{a b}{P}\) – \(\frac{R T V^{2}}{P}\) = 0 ………..(4)
When the above equation is rearranged in powers of V

V3 – [\(\frac{R T}{P}\) + b]V2 + \(\frac{a}{P}\)V – \(\frac{a b}{P}\) = 0 ……………..(5)

The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume Vc. The pressure and temperature becomes Pc and Tc respectively
V = Vc
V – Vc = 0
(V – Vc)3 = 0
V3 – 3VcV2 + 3Vc2V – Vc3 = 0 …………..(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V2, V and constant terms in (5) and (6).
-3VcV2 = –[\(\frac{R T_{c}}{P_{c}}\) + b] V2

3Vc = \(\frac{R T_{c}}{P_{c}}\) + b ………………(7)
3Vc2 = \(\frac{a}{P_{c}}\) ……………(8)
Vc3 = \(\frac{ab}{P_{c}}\) …………….(9)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 3
The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.
a = 3Vc2 Pc and
b = \(\frac{V_{c}}{3}\)

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of the moon?
Answer:
In space, there is no pressure, if we do wear a pressurized suit, our body will die. In space, we have to wear a pressurized suit, otherwise, our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurized suit (protective suits).

Question 45.
When ammonia combines with HCl, NH4Cl is formed as white dense fumes. Why do more fumes appear near HCl?
Answer:
HCl and NH4Cl molecules diffuse through the air towards each other. When they meet, they reactto
form a white powder called ammonium chloride, NH4Cl.
HCl(g) + NH3(g) ⇌ NHC1

Hydrogen chloride + ammonia ⇌ ammonium chloride.
The ring of white powder is closer to the HCl than
the NH3. This is because the NH3 molecules are lighter (smaller) and have diffùsed more quickly through the air in the tube.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 46.
A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm3. When the temperature is raised to 38°C at 1 atm does the volume of the gas Increase? If so, calculate the final volume.
Answer:
T1 = 15°C + 273;
T2 = 38 + 273
T1 = 228;
T2 = 311K
V1 = 2.58dm3;
V2 = ?
(P = 1 atom constant)
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

V2 = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T2

= \(\frac{2.58 d m^{3}}{288 K}\) × 311K
V2 = 2.78 dm3 i.e, volume increased from 2.58 dm3 to 2.78 dm3

Question 47.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen In a vessel of volume of 37.6 dm3 at 298K, and sample B is in a vessel of volume 16.5 dm3 at 298K. Calculate the number of moles in sample B.
Answer:
nA = 1.5mol; nB = ?
VA = 37.6 dm3; VB = 16.5 dm3
(T = 298K constant)
\(\frac{V_{A}}{n_{A}}=\frac{V_{B}}{n_{B}}\)

nA = \(\left[\frac{n_{A}}{n_{B}}\right] V_{B}\)

Question 48.
Sulphur hexafluoride Is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas In a steel vessel of volume 5.43 dm3 at 69.5°C, assumIng Ideal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm3
T = 69.5 + 273 = 342.5
P =?
PV = nRT
P = nRT/V
P = Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 4
P = 94.25 atm

Question 49.
Argon is an Inert gas used In light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C Is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P1 = 1.2 atm
T1 = 180°C + 273 = 291 K
T2 = 850°C + 273 = 358 K
P2 =?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P2 = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\)

P2 = \(\frac{1.2 a t m}{291 K}\) × 358 K
P2 = 1.48 atm

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 50.
A small bubble rises from the bottom of a lake where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure Is I arm. Calculate the final volume in (mL) of the bubble, If its initial volume 1.5 mL.
Answer:
T1 = 6°C + 273 = 279 K
P1 = 4 atm; V2 = 1.5 ml
T2 = 25°C + 273 = 298 K
P2 = 1 atm; V2 =?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

= \(\frac{4 a t m \times 1.5 m l \times 298 K}{279 K \times 1 a t m}\) = 6.41 mol

Question 51.
Hydrochloric acid Is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 × 10-3 dm3 of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 × 10-3 dm3
P = 742 mm of Hg
T = 298K;
m =?
m = \(\frac{P V}{R T}\)

= \(\frac{742 m m H g \times 154.4 \times 10^{-3} L}{62 m m H g L K^{-1} m o l^{-1} \times 298 K}\)

n = \(\frac{\text { Mass }}{\text { Molar Mass }}\)

Mass = n × Molar Mass
= 0.0006 × 2.016
= 0.0121 g = 12.1 mg

Question 52.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknown gas?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 5

Question 53.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure In the tanks Is 9.21 atm. calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 6
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 7

Question 54.
Combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a maximum pressure of 12 atm after which It will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal 1100 K).
Answer:
Pressure of the gas in the tank at its melting point
T = 298 K;
P1 = 2.98 atom;
T2 = 1100 K;
P2 =?
\(\frac{P_{1} P_{2}}{T_{1} T_{2}}\) =????
⇒ P2 = \(\frac{P}{T_{1}}\) × T2
= \(\frac{2.98 \text { atm }}{298 K}\) × 1100 K = 11 atm

At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

11th Chemistry Guide Gaseous State Additional Questions and Answers

I. Choose the best Answer:

Question 1.
For one mole of a gas, the ideal gas equation is ………..
(a) PV = \(\frac {1}{2}\) RT
(b) PV = RT
(c) PV = \(\frac {3}{2}\)RT
(d) PV = \(\frac {5}{2}\)RT
Answer:
(b) PV= RT

Question 2.
The SI unit of pressure is
(a) pascal
(b) atmosphere
(c) bar
(d) torr
Answer:
(a) pascal

Question 3.
Which of the following is the correct mathematical relation for Charles’ law at constant pressure?
(a) V ∝ T
(b) V ∝ t
(c) V ∝ – \(\frac {1}{T}\)
(d) all of above
Answer:
(a) V ∝ T

Question 4.
“For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature”. This statement is
(a) Boyle’s law
(b) Gay-Lussac law
(c) Avogadro’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 5.
With rise in temperature, the surface tension of a liquid …………
(a) decreases
(b) increases
(c) remaining the same
(d) none of the above
Answer:
(a) decreases

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 6.
The plot of the volume of the gas against its temperature at a given pressure is called
(a) isotone
(b) isobar
(c) isomer
(d) isotactic
Answer:
(b) isobar

Question 7.
The cleansing action of soaps and detergents is due to …………..
(a) internal friction
(b) high hydrogen bonding
(c) viscosity
(d) surface tensions
Answer:
(d) surface tensions

Question 8.
The precise value of temperature at which the volume of the gas becomes zero is
(a) -273.15 °C
(b) -273 °C
(c) -298.15°C
(d) -298 °C
Answer:
(a) -273.15 °C

Question 9.
The compressibility factor, z for an ideal gas is ………….
(a) zero
(b) less than one
(c) greater than one
(d) equal to one
Answer:
(d) equal to one

Question 10.
The value of gas constant, R, in terms of JK-1 mol-1 is
(a) 8.314
(b) 4.184
(c) 0.0821
(d) 1.987
Answer:
(a) 8.314

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 11.
Which of the following is a monoatomic gas in nature?
(a) Oxygen
(b) Hydrogen
(c) Helium
(d) Ozone
Answer:
(c) Helium

Question 12.
A mixture of gases containing 1 mole of He, 4 moles of Ne, and 5 moles of Xe. The correct order of partial pressure of the gases, if the total pressure is 10 atm is
(a) Xe < Ne < He
(b) He < Ne < Xe
(c) Xe < Ne < He
(d) He < Xe < Ne
Answer:
(b) He < Ne < Xe

Question 13.
Which one of the following is not a monoatomic gas?
(a) Neon
(b) Xenon
(c) Argon
(d) Oxygen
Answer:
(d) Oxygen

Question 14.
The process in which agas escapes from a container through a very small hole is called
(a) diffusion
(b) effusion
(c) occlusion
(d) dilution
Answer:
(a) diffusion

Question 15.
Which of the following is a tri atomic gas at room temperature?
(a) Oxygen
(b) Helium
(c) Ozone
(d) Nitrogen
Answer:
(c) Ozone

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 16.
An unknown gas X diffuses at a rate of 2 times of oxygen at the same temperature and pressure. The molar mass (in g mol-1) of the gas ‘X is (Molar mass of oxygen is 32 g mol-1)
(a) 8
(b) 16
(c) 20
(d)12
Answer:
(a) 8

Question 17.
Among the following, which is deadly poison?
(a) CO2
(b) HCN
(c) HCl
(d) NH3
Answer:
(b) HCN

Question 18.
The gases which deviate from ideal behavior at
(a) low temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) high temperature and high pressure
Answer:
(b) high temperature and low pressure

Question 19.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called temperature
(a) inversion
(b) ideal
(c) Boyle
(d) reversible
Answer:
(c) Boyle

Question 20.
The pressure of a gas is equal to ………..
(a) \(\frac {F}{a}\)
(b) F x a
(c) \(\frac {a}{F}\)
(d) F – a
Answer:
(a) \(\frac {F}{a}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 21.
The pressure correction introduced by Van der Waals is,
Pideal =
(a) P + \(\frac{a V^{2}}{n^{2}}\)

(b) P + \(\frac{a n}{V^{2}}\)

(c) P + \(\frac{a n^{2}}{V^{2}}\)

(d) P + \(\frac{a V}{n^{2}}\)
Answer:
(c) P + \(\frac{a n^{2}}{V^{2}}\)

Question 22.
Statement-I: The pressure cooker takes more time for cooking at high altitude.
Statement-II: Air is subjected to Earth’s gravitational force. The pressure of air gradually decreases from the surface of the Earth to higher altitude.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(e) Statement-I is wrong but Statement-II is correct
(ð) Statement-I is correct but Statement-II is wrong
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 23.
Van der waals equation for one mole of a gas is
(a) (P + \(\frac{a}{V}\))(V-b) = RT
(b) (P – \(\frac{a}{V^{2}}\))(V + -b) = RT
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT
(d) (P + \(\frac{a}{V}\))(V + b) = RT
Answer:
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT

Question 24.
The standard atmospheric pressure at sea level at 0°C is equal to ……………..
(a) 1 mm Hg
(b) 76 mm Hg
(c) 760 mm Hg
(d) 680 mm Hg
Answer:
(c) 760 mm Hg

Question 25.
The unit of Van der waals constant ‘b’ is
(a) lit mol-1
(b) lit mol
(c) atm lit mol-1
(d) atm lif-1 mol-2
Answer:
(a) lit mol-1

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 26.
The temperature above which a gas cannot be liquefied even at high pressure is called is ________ temperature.
(a) ideal
(b) inversion
(c) critical
(d) real
Answer:
(c) critical

Question 27.
Which one of the following represents Charles’ law?
(a) PV = Constant
(b) \(\frac {V}{T}\) = Constant
(c) VT Constant
(d) \(\frac {T}{V}\) = R
Answer:
(b) \(\frac {V}{T}\) = Constant

Question 28.
Which of the following gas has the highest critical temperature?
(a) NH3
(b) CO2
(c) N2
(d) CH4
Answer:
(a) NH3

Question 29.
\(\frac {P}{T}\) = Constant is known as …………..
(a) Boyle’s law
(b) Charles’ law
(c) Gay Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay Lussac’s law

Question 30.
The temperature below which a gas obeys the Joule-Thomson effect is called ______ temperature.
(a) critical
(b) Inversion
(c) ideal
(d) real
Answer:
(b) Inversion

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 31.
The value of the Universal gas constant in an ideal gas equation is equal to ………….
(a) 8.3 14 KJ
(b) 0.082057 dm3 atm mol-1 K-1
(c) 1 Pascal
(d) 8.314 x 10-2Pascal
Answer:
(b) 0.082057 dm3 atm mol-1 K-1

Question 32.
A sample of a given mass of gas at constant temperature occupies a volume of 95 cm3 under pressure of 10.13 × 104 Nm-2. At the same temperature, its volume at a pressure of 10.13 × 104 Vm-2 is
(a) 190 cm3
(b) 93 cm3
(c) 46.5 cm3
(d) 4.75 cm3
Answer:
(b) 93 cm3

Question 33.
Which law is used in the isotopic separation of deuterium and protium?
(a) Boyle’s law
(b) Charles’ law
(c) Graham’s law
(d) Gay Lussac’s law
Answer:
(c) Graham’s law

Question 34.
The use of hot air balloons in sports and meteorological observations is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 35.
The value of critical volume is equal in terms of Vander Waals constant is ……….
(a) 3b
(b) \(\frac{8a}{27 Rb}\)
(c) \(\frac{a}{27 b^{2}}\)
(d) \(\frac{2a}{Rb}\)
Answer:
(a) 3b

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 36.
7.0 g of a gas at 300K and 1 atm occupies a volume of 4.1 litre. What is the molecular mass of the gas?
(a) 42
(b) 38.24
(c) 14.5
(d) 46.5
Answer:
(a) 42

Question 37.
If most probable velocity is represented by a and fraction possessing it by / then with increase in temperature which one of the following is correct?
(a) α increases, f decreases
(b) α decreases, f increases
(c) Both α and f decrease
(d) Both α and f increase
Answer:
(a) α increases, f decreases

Question 38.
The value of critical pressure of CO2 is ……………….
(a) 173 atm
(b) 73 atm
(c) 1 atm
(d) 22.4 atm
Answer:
(b) 73 atm

Question 39.
To which of the following gaseous mixtures Dalton’s law is not applicable?
(a) Ne + He + SO2
(b) NH3 + HCl + HBr
(c) O2 + N2 +CO2
(d) N2 + H2 + O2
Answer:
(b) NH3 + HCl + HBr

Question 40.
The substance used in the adiabatic process of liquefaction is ……………
(a) liquid helium
(b) gadolinium sulphate
(c) iron sulphate
(d) liquid ammonia
Answer:
(b) Gadolinium sulphate

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 41.
50 mL of gas A effuses through a pinhole in 146 seconds. The same volume of CO2 under identical condition effuses in 115 seconds. The molar mass of A is
(a) 44
(b) 35.5
(c) 71
(d) None of these
Answer:
(c) 71

Question 42.
The molecules of gas A travel four times faster than the molecules of gas B at the same temperature. The ratio of molecular weight MA/ MB is ………….
(a) 1/16
(b) 4
(c) 1/4
(d) 16
Answer:
(a) 1/16

Question 43.
Which of the following gases is expected to have the largest value of van der Waals constant ‘a’?
(a) He
(b) H2
(c) NH3
(d) O2
Answer:
(c) NH3

Question 44.
In van der Waals equation of state for a real gas, the term that accounts for intermolecular forces is
(a) Vm – b
(b) P + \(\frac{a}{V m^{2}}\)
(c) RT
(d) \(\frac{1}{R T}\)
Answer:
(b) P + \(\frac{a}{V m^{2}}\)

Question 45.
The value of Vander Waals constant “a” is maximum for ……………….
(a) helium
(b) nitrogen
(c) methane
(d) ammonia
Answer:
(d) Ammonia

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 46.
Which of the following has a non-linear relationship?
(a) P vs V
(b) P vs \(\frac{1}{V}\)
(c) both (a) and (b)
(d) none of these
Answer:
(a) P vs V

Question 47.
Why is that the gases show ideal behavior when the volume occupied is large?
(a) So that the volume of the molecules can be neglected in comparison to it.
(b) So that the pressure is very high
(c) So that the Boyle temperature of the gas is constant
(d) all of these
Answer:
(a) So that the volume of the molecules can be neglected in comparison to it.

Question 48.
At a given temperature, the pressure of a gas obeying the Van der Waals equation is
(a) less than that of an ideal gas
(b) more than that of an ideal gas
(c) more or less depending on the nature of gas
(d) equal to that of an ideal gas
Answer:
(a) less than that of an ideal gas

Question 49.
The rate of diffusion of a gas is ………….
(a) directly proportional to its density
(b) directly proportional to its molecular mass
(c) directly proportional to the square root of its molecular mass
(d) inversely proportional to the square root of its molecular mass
Answer:
(d) inversely proportional to the square root of its molecular mass

Question 50.
Maximum deviation from ideal gas is expected from
(a) CH4(g)
(b) NH3(g)
(c) H2(g)
(d) N2(g)
Answer:
(b) NH3(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

II. Very short question and answers (2 Marks):

Question 1.
Identify the elements that are in the gaseous state under normal atmospheric conditions.
Answer:

  • Hydrogen, nitrogen, oxygen, fluorine, and chlorine exist as gaseous diatomic molecules.
  • Another form of oxygen namely the ozone triatomic molecule exists as a gas at room temperature.
  • Noble gases namely helium, neon, argon, krypton, xenon, and radon are monoatomic gases.

Question 2.
State Gay Lussac ‘s law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T or \(\frac{P}{T}\) = constant K
If P1 and P2 are the pressures at temperatures T1 and T2, respectively, then from Gay Lussac’s law
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 3.
Define pressure. Give its units.
Answer:

  • Pressure is defined as the force exerted by a gas on unit area of the wall. Force F
  • Pressure = \(\frac {Force}{Area}\) = \(\frac {F}{a}\)
  • The SI unit of pressure is Pascal (Pa)

Question 4.
What is the Compressibility factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compressibility factor. Mathematically,
Z = \(\frac{P V}{n R T}\)

Question 5.
Deep-sea divers ascend slowly and breathe continuously by the time they reach the surface. Give reason.
Answer:

  • For every 10 m of depth, a diver experiences an additional 1 atm of pressure due to the weight of water surrounding him.
  • At 20 m, the diver experiences a total pressure of 3 atm. So the most important rule in diving is never to hold a breath.
  • Divers must ascend slowly and breathe continuously allowing the regulator to bring the air pressure in their lungs to 1 atm by the time they reach the surface.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 6.
How does cooling is produced in the Adiabatic Process?
Answer:
In the Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10-4 K i.e., as low as 0 K can be achieved.

Question 7.
What is the reason behind the cause of ear pain while climbing a mountain? How it can be rectified?
Answer:

  • When one ascends a mountain in a plain, the external pressure drops while the pressure within the air cavities remains the same. This creates an imbalance.
  • The greater internal pressure forces the eardrum to bulge outward causing pain.
  • With time and with the help of a yawn or two, the excess air within your ear’s cavities escapes thereby equalizing the internal and external pressure and relieving the pain.

Question 8.
Write a note on the Consequence of Boyle’s law.
Answer:
The pressure-density relationship can be derived from the Boyle’s law as shown below.
P1V1 = P2V2
P1 \(\frac{m}{d_{1}}\) = P2 \(\frac{m}{d_{2}}\)
where “m” is the mass, d1 and d2 are the densities of gases at pressure P1 and P2.
\(\frac{P_{1}}{d_{1}}=\frac{P_{2}}{d_{2}}\)
In other words, the density of a gas is directly proportional to pressure.

Question 9.
What are the applications of Charles’ law?
Answer:

  • The hot air inside the balloon rises because of its decreased density and causes the balloon to float inside the balloon rises because of its decreased density and causes the balloon to float.
  • If you take a helium balloon outside on a chilly day, the balloon will crumble. Once you get back into a warm area, the balloon will return to its original shape. This is because, in accordance with Charles’ law, a gas like helium takes up more space when it is warm.

Question 10.
Write a note on the application of Dalton’s law.
Answer:
In a reaction involving the collection of gas by downward displacement of water, the pressure of dry vapor collected can be calculated using Dalton’s law.
Pdry gas collected = Ptotal – Pwatervapour
Pwatervapour has generally referred to as aqueous tension and its values are available for air at various temperatures.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 11.
Define Dalton’s law of partial pressure.
Answer:
Dalton’s law of partial pressure:
It states that the total pressure of a mixture of gases is the sum of partial pressures of the gases present.
Ptotal = P1 + P2 + P3

Question 12.
What happens when a balloon is moved from an ice cold water bath to a boiling water bath?
Answer:
If a balloon is moved from an ice-cold water bath to a boiling water bath, the temperature of the gas increases. As a result, the gas molecules inside the balloon move faster and gas expands. Hence, the volume increases.

Question 13.
Write notes on the coefficient of expansion(α).
Answer:
The relative increase in volume per °C (α) is equal to \(\frac{V}{V_{0} T}\)
Therefore, \(\frac{V}{V_{0} T}\)
TV = V0(αT + 1)
Charles found that the coefficient of expansion is approximately equal to 1/273. It means that at constant temperature for a given mass, for each degree rise in temperature, all gases expand by 1/273 of their volume at 0°C.

Question 14.
Define Graham’s law of diffusion.
Answer:
Graham’s law of diffusion:
The rate of diffusion or effusion is inversely proportional to the square root of the molecular mass of a gas through an orifice.
\(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}}\)
rArB = rate of diffusion of gases A, B
MA, MB = Molecular mass of gases A, B

Question 15.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 16.
Explain the applications of Graham’s law of diffusion.
Answer:

  • Graham’s law of diffusion is useful to determine the molecular mass of the gas if the rate of diffusion is known.
  • Graham’s law forms the basis of the process of enriching the isotopes of U235 from other isotopes and also useful in the isotopic separation of deuterium and protium.

Question 17.
State Graham’s law of diffusion.
Answer:
The rate of diffusion or effusion is inversely proportional to the square root of molar mass. This statement is called Graham’s law of diffusion/effusion.
Mathematically, rate of diffusion ∝ \(\frac{1}{M}\)

Question 18.
What is Boyle temperature?
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

III. Short Question and Answers (3 Marks):

Question 1.
Write notes on Boyle’s point.
Answer:
The temperature at which a real gas obeys the ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point. The Boyle point varies with the nature of the gas. Above the Boyle point, for real gases, Z > 1, i.e., the real gases show a positive deviation. Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum, and then increase with the increase in pressure.

Question 2.
What are the consequences of Boyle’s law?
Answer:
1. if the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
2. Boyle’s law also helps to relate the pressure to density.
P1V1 = P3V3 (Boyle’s law)
\(P_{1} \frac{m}{d_{1}}=P_{2} \frac{m}{d_{2}}\)
Where ‘m’ is the mass, d1 and d2 are the densities of gases at pressure P1 and P2. The density of the gas is directly proportional to pressure.

Question 3.
48 liter of dry N2 is passed through 36g of H2O at 27°C and this results In a loss of 1.20 g of water. Find the vapour pressure of water?
Answer:
Water loss is observed because of the escape of water molecules with N2 gas. These water vapour occupy the volume of N2 gas i.e., 48 litres.
Using,
PV = \(\frac{m}{M}\)RT;
V = 48L
m = 1.2g;
M = 18u;
R = 0.082 L atm K-1mol-1
T = 27 + 273 = 300 K
P = \(\frac{1.2}{18} \times \frac{0.0821 \times 300}{48}\) = 0.034 atm

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 4.
A flask of capacity one litre is heated from 25°C to 35°C. What volume of air will escape from the flask?
Answer:
Applying, \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V1 = 1L;
T1 = 25 + 273 = 298 K
V2 = ?
T2 = 35 + 273 = 308 K
V2 = \(\frac{1}{298}\) × 308 = 1.033 L
Capacity of Flask = 1 L
So, volume of air escaped = 1.033 – 1 = 0.033 L = 33 mL

Question 5.
Discuss the graphical representation of Boyle’s law.
Answer:
Boyle’s law is applicable to all gases regardless of their chemical identity (provided the pressure is low). Therefore, for a given mass of a gas under two different sets of conditions at constant temperature we can write,
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 10
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 11

Question 6.
A certain gas takes three times as long to effuse out as helium. Find its molecular mass.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 12

Question 7.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant. Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.
Boyle’s law: V ∝ \(\frac{1}{P}\)
Charles’ law: V ∝ T
Avogadro’s law: V ∝ n
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
V ∝ \(\frac{nT}{P}\)
V = \(\frac{nRT}{P}\)
where R = Proportionately constant.
The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where R is also known as Universal gas constant.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 8.
Write notes on the compressibility factor for real gases.
Answer:
The compressibility factor Z for real gases can be rewritten,
Z = PVreal / nRT …………..(1)
Videal = \(\frac{n R T}{P}\) ……………(2)
substituting (2) in (1)
where Vreal is the molar volume of the real gas and Videal is the molar volume of it when it behaves ideally.

Question 9.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V, and n and their relationships were governed by the gas laws studied so far.
Boyle’s law V ∝ P1
Charles law V ∝ T
Avogadro’s law V ∝ n
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
V ∝ \(\frac{n T}{P}\)

V = \(\frac{n R T}{P}\)
where, R is the proportionality constant called universal gas constant.
The above equation can be rearranged to give the ideal gas equation
PV = nRT

Question 10.
Explain the different methods used for the liquefaction of gases.
Answer:

  • Linde’s method: Joule-Thomson effect is used to get liquid air or any other gas.
  • Claude’s process: In addition to the Joule-Thomson effect, the gas is allowed to perform mechanical work so that more cooling is produced.
  • Adiabatic process: This method of cooling is produced by removing the magnetic property of magnetic material e.g. Gadolinium sulphate. By this method, a temperature of 10-4 K i.e. as low as zero Kelvin can be achieved.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

IV. Long Question and Answers (5 Marks):

Question 1.
Derive Van der Waals equation of state.
Answer:
J.D.Van der Waals made the first mathematical analysis of real gases. His treatment provides us an interpretation of real gas behaviour at the molecular level. He modified the ideal gas equation PV = nRT by introducing two correction factors, namely, pressure correction and volume correction.

Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.

Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
P’ ∝ p2
p = \(\frac{n}{v}\)
where n is the number of moles of gas and V is the volume of the container
⇒ P’ = \(\frac{n^{2}}{V^{2}}\)
⇒ P’ = a\(\frac{n^{2}}{V^{2}}\)

where a is the proportionality constant and depends on the nature of gas.
Therefore, ideal P = P + a\(\frac{n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V’ to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\)π(2r)3
= \(\left|\frac{-8}{(3 \pi)}\right|\) = 8Vm

where Vm is a volume of a single molecule
Excluded volume for single-molecule = \(\frac{8 V_{m}}{2}\) = 4Vm

Excluded volume for n molecule = n(4Vm) = nb
Where b is van der Waals constant which is equal to 4 Vm
=> V’ = nb
Videal = V nb

Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 2.
Derive the relationship between Van der Waals constants and critical constants.
Answer:
The van der Waals equation for n moles is
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
For 1 mole
(p + \(\frac{a}{V^{2}}\))(V – b) = RT
From the equation, we can derive the values of critical constants Pc, Vc, and Tc in terms of a and b, the van der Waals constants, On expanding the above equation
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0

Multiply the above equation by
\(\frac{V_{2}}{P}\)(PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V3 + \(\frac{a V}{P}\) + -bV2 – \(\frac{a b}{p}\) – \(\frac{R T V^{2}}{P}\) = 0

when the above equation is rearranged in powers of V

V3 \(\frac{R T}{P}\) + bV2 \(\frac{a}{P}\) V – \(\frac{a b}{P}\) = 0.

The above equation is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point, all these three solutions of Vc are equal to the critical volume. The pressure and temperature becomes Pc and Tc respectively
i.e., V = Vc
V – Vc = 0
V – (Vc)3 = 0
V3 – 3VcV2 + 3VVc2 – Vc3 = 0.
As equation identical with the equation above, we can equate the coefficients of Vc, V, and constant terms.

-3VcV2 = –\(\frac{R T_{C}}{P}\) + bV2

3Vc = \(\frac{R T_{C}}{P_{C}}\) + b ……….(1)

3Vc2 = \(\frac{a}{P_{C}}\) …………..(2)

Vc3 = \(\frac{a b}{P_{C}}\) …………..(3)

Divide equation (3) by equation (2)

\(\frac{V_{c}^{3}}{3 V_{C}^{2}}=\frac{a b / P_{C}}{a / P_{C}}\)

\(\frac{V_{C}}{3}\) = b
i.e., Vc = 3b ……………(4)

when equation (4) is substituted in (2)
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 13
The critical constants can be calculated using the values of Vander walls constant of a gas and vice versa.

a = 3Vc2 Pc and b = \(\frac{V_{C}}{3}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 3.
Explain Andrew’s isotherm for Carbon dioxide.
Answer:
Thomas Andrew gave the first complete data on the pressure-volume temperature of a substance in the gaseous and liquid states. He plotted isotherms of carbon dioxide at different temperatures which are shown in Figure. From the plots we can infer the following.

At low-temperature isotherms, for example, at 130°C as the pressure increases, the volume decreases along with AB and is a gas until point B is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist and the pressure remains constant. At C, the gas is completely converted into liquid. If the pressure is higher than at C, only the liquid is compressed so, there is no significant change in the volume. The successive isotherms show a similar trend with the shorter flat region. i.e., The volume range in which the liquid and gas coexist becomes shorter.

At the temperature of 31.1°C the length of the shorter portion is reduced to zero at point P. In other words, the CO2 gas is liquefied completely at this point. This temperature is known as the liquefaction temperature or critical temperature of CO3. At this point, the pressure is 73 atm. Above this temperature, CO3 remains as a gas at all pressure values. It is then proved that many real gases behave in a similar manner to carbon dioxide.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 14

Question 4.
Explain Boyle’s law experiment:
Answer:
Robert Boyle performed a series of experiments to j study the relation between the pressure and volume of gases. The schematic diagram of the apparatus j used Boyle is shown in the figure.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 15

Mercury was added through the open end of the apparatus such that the mercury level on both ends is equal as shown in figure (a). Add more amount of mercury until the volume of the trapped air is reduced to half of its original volume as shown in figure (b). The pressure exerted on the gas by the addition of excess mercury is given by the difference in mercury levels of the tube.

Initially, the pressure exerted by the gas is equal to 1 atm as the difference in height of the mercury levels is zero. When the volume is reduced to half, the difference in mercury levels increases to 760 mm. Now the pressure exerted by the gas is equal to 2 atm. It led him to conclude that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, Boyle’s law can be written as
V ∝ \(\frac{1}{P}\) ……….(1)
(T and n are fixed, T-temperature, n- number of moles)
V = k × \(\frac{1}{P}\) ……….(2)
k – proportionality constant When we rearrange equation (2)
PV = k at constant temperature and mass.

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