250+ TOP MCQs on States of Matter – Behaviour of Real Gases: Deviation from Ideal Gas Behaviour and Answers

Chemistry Multiple Choice Questions on “States of Matter – Behaviour of Real Gases: Deviation from Ideal Gas Behaviour”.

1. The plot PV vs v at constant temperature is a straight line for real gases.
a) true
b) false
Answer: b
Clarification: The plot of PV vs P is not a straight line for real gases because they deviate from Ideal behaviour. are there are two types of deviations one is a positive deviation and the other is a negative deviation.

2. PV/nRT is known as ____________
a) compressibility factor
b) volume factor
c) pressure factor
d) temperature factor
Answer: a
Clarification: PV/nRT is known as compressibility factor and is represented by the letter Z. It is a ratio of PV and nRT; where p is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is temperature.

3. Which of the following conditions do you think a real gas behaves as an ideal gas?
a) high pressure
b) low pressure
c) intermediate pressure
d) at any pressure
Answer: b
Clarification: At low-pressure conditions, Z = 1 handset behaves as an ideal gas but at high-pressure Z is greater than 1 and for intermediate pressure that is less than 1. So at low-pressure condition, a real gas behaves as an ideal gas.

4. What is the temperature known as where a real gas obeys Boyle’s law or as an ideal gas?
a) Boyle temperature
b) Charge temperature
c) Critical temperature
d) Absolute Temperature
Answer: a
Clarification: The temperature at which a real gas obeys Boyle’s law and other ideal gas law at a certain range of pressure is called Boyle temperature or Boyle point. It is unique for every gas and depends upon its nature.

5. Compressibility can be expressed as _______
a) real volume divided by the ideal volume
b) real universal gas constant by ideal universal gas constant
c) real temperature by ideal temperature
d) real volume divided by real pressure
Answer: a
Clarification: The deviation of real gas behaviour from ideal gas behaviour is known from the compressibility factor. This compressibility factor can also be measured as the ratio of real volume to ideal volume.

6. Above Boyle temperature real gases show __________ deviation from ideal gases.
a) positive
b) negative
c) no
d) both positive and negative
Answer: a
Clarification: Above Boyle temperature, the value of the compressibility factor is greater than 1. So the gases show positive deviation from ideal gases as the forces of attraction between the gas molecules are very low.

7. Which of the following is a corrected equation of ideal gas equation?
a) (P – an²V²)(V – nb) = nRT
b) (P – an²/V²)(V + nb) = nRT
c) (P + an²/V²)(V – nb) = nRT
d) (P – an²/V²)(V – nb) = nRT
Answer: d
Clarification: (P – an²/V²)(V – nb) = nRT; where p is pressure, a is the magnitude of intermolecular attractive forces within a gas, n is the number of moles, v is volume, b is a van der Waal constant, R is the universal gas constant and T is temperature.

8. The value of a in van der Waal equation is _____________ /dependent on ___________
a) pressure
b) temperature
c) pressure and temperature
d) independent of pressure and temperature
Answer: d
Clarification: Value of an in van der Waal equation represents a measure of the magnitude of intermolecular attractive forces within the gas and it is also independent of temperature and pressure. The van der Waal’s equation is given by (P – an²/V²)(V – nb) = nRT.

9. What are the units of “b” in van der Waals equation?
a) L/mol
b) L mol
c) 1/L mol
d) L
Answer: a
Clarification: 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.

10. 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
Answer: c
Clarification: Compressibility factor Z = PV/nRT; Z = 50 atm x (500/1000) ml / 2 x 0.082 x 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.