Grand Chemistry MCQs Test of Gaseous State for Class 11 | Interactive Quiz

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🔥🌟 Grand Chemistry MCQs Test of Gaseous State for Class 11 | Interactive Quiz 🧪💡| Learn & Practice!

1. According to Graham’s Law of diffusion, the ratio of diffusion of H₂ and O₂ are respectively:

a) 1:2

b) 2:1

c) 1:4

d) 4:1

Correct Answer: (d)

r₁/r₂= √M₂/√M₁ ⇒ r₁/r₂ = √32/√2 ⇒ r₁/r₂= √16 ⇒ r₁/r₂= 4:1

2. Collection of gas over water is an example of:

a) Graham’s law

b) Gay-Lussac law

c) Dalton’s law

d) Avogadro’s law

Correct Answer: (c)

Collection of gas over water is an example of Dalton’s law. The pressure of dry gas is calculated by using Dalton’s law. Pdry gas = Pmoist gas – Pwater vapours.

3. The molar volume of oxygen gas is maximum at:

a) 0°C and 1 atm

b) 0°C and 2 atm

c) 25°C and 1 atm

d) 25°C and 2 atm

Correct Answer: (c)

The molar volume of a gas varies directly with temperature and varies inversely with pressure. The molar volume of a gas is maximum at highest temperature and least pressure. In option c and d, temperature is highest hence deciding factor will be pressure which is least in option c.

4. The volume of gas would be theoretically zero at:

a) 0°C

b) 273 K

c) -273°C

d) 0 K

Correct Answer: (d)

The volume of gas would be theoretically zero at 0 K (-273°C).

5. If the Kelvin temperature of ideal gas is increase to double and pressure is reduced to one half, the volume of gas will:

a) Reduced to half

b) Four times

c) Double

d) Remains same

Correct Answer: (b)

PV=nRT ⇒ PV = T (R and n = constant) ⇒ V = T/P ⇒ V = 2/ ½ = 2 x 2 = 4

Thus, doubling the Kelvin temperature and halving the pressure quadruples the volume.

6. The molar volume of oxygen gas is 22.4 dm³ at:

a) 25 K & 0.5 atm

b) 0 K and 1 atm

c) 25°C and 0.5 atm

d) 0°C and 1 atm

Correct Answer: (d)

The molar volume of a gas is 22.4 dm³ at STP i.e. at 0°C and 1 atm.

7. Under similar condition CH₄ gas diffuses........ times faster than SO₂ gas:

a) 1.5 times

b) 4 times

c) 16 times

d) 2 times

Correct Answer: (d)

r₁/r₂ = √M₂/√M₁ ⇒ r₁/r₂ = √64/√16 ⇒ r₁/r₂ = √4 ⇒ r₁/r₂ = 2

8. Which one of the following statement is INCORRECT about the gas molecules?

a) Their collision is elastic

b) They have large spaces

c) Their molar mass depends upon temperature

d) They possess kinetic energy

Correct Answer: (c)

The molar mass of gas is independent of temperature.

9. The diffusion rate of C₃H₈ and CO₂ are same because:

a) Both contains carbon atoms

b) Both have same molar mass

c) Both are denser than air

d) Both are poly atomic gases

Correct Answer: (b)

The rate of diffusion depends upon molar masses of gases. The gases with same molar masses would have same rate of diffusion. The given gases ethane (C₃H₈) and CO₂ have same molar mass of 44 gmol⁻¹.

10. Real gas reaches the ideal behavior at:

a) High temperature and high pressure

b) Low temperature and high pressure

c) High temperature and low pressure

d) Low temperature and low pressure

Correct Answer: (c)

High temperature and low pressure make the ideal.

11. If the ratio of the rates of diffusion of the two gases A and B is 4:1, the ratio their densities is:

a) 1:4

b) 1:8

c) 1:2

d) 1:16

Correct Answer: (d)

Using Graham’s law, the ratio their densities is 1:16.

12. Relative rates of diffusions of two gases X and Y are found to be 3:2. If the density of Y is 27 the density of X is:

a) 27

b) 54

c) 4

d) 12

Correct Answer: (d)

Using Graham’s law, the density of X is 12.

13. 1 liter of a gas weighs 2 g at 300 K and 1 atm pressure. If the pressure is made 75 atm, at which of the following temperatures will 1 L of the same gas weigh 1 g?

a) 450 K

b) 800 K

c) 600 K

d) 900 K

Correct Answer: (a)

The correct answer is 450 K. PV = nRT = m/MRT 

For same gas 

P₁V₁ = n₁RT₁ = m₁/MRT₁ 

P₂V₂ = n₂RT₂ = m₂/MRT₂ 

P₁V₁/ P₂V₂ = n₁RT₁/ n₂RT₂ = m₁/MRT₁/ m₂/MRT₂ 

P₁V₁/ P₂V₂ = m₁RT₁/ m₂RT₂ 

 P₁V₁/ P₂V₂ = m₁T₁/ m₂T₂ 

 1x1/ 0.75x1 = 2x300/ 1x T₂ = 450K

14. What are the units of “b” in van der Waals equation?

a) L

b) 1/L mol

c) L mol

d) L/mol

Correct Answer: (d)

The ideal gas equation is given as (P – an²/V²)(V – nb) = nRT. So by considering the equation, we can understand that the units of the volume are equal to the units of a number of moles X be so the units of b. So b’s units = volume /number of moles, so it is L/mol.

15. A gas that is of 2 moles occupies a volume of about 500 ml at 300 Kelvin and 50 atmospheric pressure, calculate the compressibility factor of the gas.

a) 1.863

b) 0.7357

c) 0.5081

d) 1.8754

Correct Answer: (c)

Compressibility factor (Z) = PV/nRT; Z = 50 atm × (500/1000) ml / 2 × 0.082 × 300 k = 25/6 × 8.2 = 0.5081. That means Z < 1, so this is a negative deviation from ideal gas behaviour. So the gas is more compressible than expected.

16. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is

a) nb

b) an²/V²

c) – an²/V²

d) -nb

Correct Answer: (b)

The term that corrects for the attractive forces present in a real gas in the van der Waal's equation is an²/V²

17. An ideal gas expands according to PV = constant. On expansion the temperature of gas

a) will rise

b) will remain constant

c) Will drop

d) Cannot be determined

Correct Answer: (b)

Since gas follows PV = constant (Boyle’s law), temperature must be held constant.

18. The temperature of a gas is 100 K, it is heated until it is 200 k then, what do you understand regarding kinetic energy in this process?

a) Halved

b) Tripled

c) Quadrupled

d) Doubled

Correct Answer: (d)

According to the kinetic molecular theory of gases, if the Absolute Temperature of a gas is doubled then the kinetic energy is also doubled because they are directly proportional to each other so here the answer is doubled. If the temperature doubles from 100 K to 200 K, kinetic energy also doubles.

19. The value of ‘b’ for carbon dioxide is given as 42.69 × 10⁻⁶ m³/mol. What do you think is the volume of a molecule?

a) 7.59 m³

b) 7.03 m³

c) 76.09 m³

d) 7.09 m³

Correct Answer: (d)

From van der Waal’s equation (P + an²/V²)(V – nb) = nRT, 

we know that V = b/Nᴀ = 42.69 × 10⁻⁶ m³/ mol/ 6.023 x 10²³ molecules/mol. That equals 7.09 m³/molecule. So the volume of a molecule is 7.09 m³.

20. What is the corresponding pressure 32.8 psi in kpa?

a) 226.085 kPa

b) 2.23 kPa

c) 1695.8 kPa

d) None of these

Correct Answer: (a)

14.7 psi = 101.325 kPa 

 1 psi = 101.325/14.7 kPa (6.892) 

To convert psi to kPa, multiply the pressure by conversion factor 10.1325/14.7 (6.892) 

32.8 psi = 10.1325/14.7 × 32.8 = 226.085 kPa

21. When the pressure approaches zero, all gases reaches a value of z ………..

a) less than 1

b) greater than 1

c) equal to 1

d) none of them

Correct Answer: (c)

When the pressure approaches zero, all gases reaches a value of z equal to 1.

22. Out of the following four molecules, which molecule does NOT possess London forces?

a) He

b) Ne

c) H₂

d) NH₃

Correct Answer: (d)

London forces are only present in non-polar molecules and in inert gases. NH₃ is a polar molecule and has hydrogen bond rather than London forces.

23. At the same temperature and pressure two identical balloons were filled with methane and sulphur dioxide. If sulphur dioxide escaped at the rate of 100 ml/sec, the methane would escape at the rate of:

a) 50 ml/s

b) 100 ml/s

c) 200 ml/s

d) 400 ml/s

Correct Answer: (c)

Using Graham’s law of diffusion, the rate of escape of methane is found to be 200 ml/s.

24. Which property is same for both normal hydrogen and deuterium?

a) Boiling point

b) Freezing point

c) Bond energy

d) Bond length

Correct Answer: (d)

Bond length depends upon the electrostatic force of attraction. The two isotopes have same atomic number therefore inter nuclear distance is same. Substituting a different isotope in the atom should not change the bond length. Since, the electrical environment around the isotope is identical.

25. The molar volume of CO₂ is maximum at:

a) STP

b) 127ºC and 1 atm

c) 0ºC and 2 atm

d) 273ºC & 2 atm

Correct Answer: (b)

The molar volume of a gas is maximum at high temperature (direct relation) and low pressure (inverse relation). In option b, pressure is lowest. 

 PV = nRT ⇒ V = nRT/V ⇒ V = RT/P (for n = 1 mole) 

 Molar volume of CO₂ at STP (273K and 1 atm) = V 

= 0.0821 x 273/1 = 22.4 L 

 Molar volume of CO₂ at 127ºC (400K) and 1 atm = V 

= 0.0821 x 400/1 = 32.84 L 

 Molar volume of CO₂ at 0ºC (273K) and 2 atm = V 

= 0.0821 x 273/2 = 11.2 L 

 Molar volume of CO₂ at 273ºC (546 K) and 2 atm = V 

= 0.0821 x 546/2 = 22.4 L

26. Which one of the following equation is used to determine the density of a gas?

a) d = MP/RT

b) d = RT/ MP

c) d = MRT

d) d = RP/MT

Correct Answer: (a)

The density expression; d = PM/RT

27. If absolute temperature of a gas is doubled and the pressure is reduced to one half, the volume of the gas will

a) Be doubled

b) Reduce to ¼

c) Remain unchanged

d) Quadruple

Correct Answer: (d)

According to Charle’s law, doubling the absolute temperature doubles the volume. According to Boyle’s law, reducing the pressure to one half doubles the volume (inverse relation). Now adding two volumes i.e. 2 + 2 = 4 volumes. Hence under these conditions, the volume of gas will be quadrupled i.e. 4 times.

28. The order of the rate of diffusion of gases NH₃, SO₂, Cl₂ and CO₂ is:

a) NH₃>SO₂>Cl₂ >CO₂

b) NH₃ > CO₂ > SO₂ > Cl₂

c) NH₃ > CO₂ > Cl₂ > SO₂

d) Cl₂>SO₂>CO₂>NH₃

Correct Answer: (b)

The rate of diffusion of gases varies inversely with their molar masses. To find out the order of the diffusion rate, arrange the gases in increasing order of their molar masses. This order will be the decreasing order of diffusion rate of gases. 

 Order of increasing molar masses ………. NH₃ (17) > CO₂ (44) > SO₂ (64) > Cl₂ (71) 

 Order of decreasing diffusion rate ……… NH₃ > CO₂ > SO₂ > Cl₂

29. The value of universal gas constant is/are:

a) 8.3143 J-mol⁻¹-K⁻¹

b) 0.0821 atm-dm³-mol⁻¹-K⁻¹

c) 1.987 cal-mol⁻¹-K⁻¹

d) All of them

Correct Answer: (d)

SI value of R ………….8.3143 J-mol⁻¹-K⁻¹ 

Non-SI value of R …. 0821 atm-dm³-mol⁻¹-K⁻¹ 

Caloric value of R … 1.987 cal-mol⁻¹-K⁻¹

30. The compressibility factor for an ideal gas is

a) 1.5

b) 1

c) -1

d) ∞

Correct Answer: (b)

The compressibility factor for an ideal gas is 1.

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