Q: For an adiabatic process involving an ideal gas, the pressure and volume are related by:

PV^\gamma = \text{constant}. Combining the first law with Q = 0 and the ideal gas equation gives PV^\gamma = constant, where \gamma = \dfrac{C_p}{C_v}.

Q: For an isothermal process involving a fixed mass of an ideal gas:

PV = \text{constant}. At constant T, PV = nRT makes the product PV a constant — this is Boyle's law.

Q: For an adiabatic process, the relation between temperature and volume is:

TV^{\gamma - 1} = \text{constant}. Substituting P = \dfrac{nRT}{V} into PV^\gamma = constant gives TV^{\gamma-1} = constant.

Question 1

For an adiabatic process involving an ideal gas, the pressure and volume are related by:

Accepted Answer

PV^\gamma = 	ext{constant}. Combining the first law with Q = 0 and the ideal gas equation gives PV^\gamma =  constant, where \gamma = \dfrac{C_p}{C_v}.

Question 2

For an isothermal process involving a fixed mass of an ideal gas:

Accepted Answer

PV = 	ext{constant}. At constant T, PV = nRT makes the product PV a constant — this is Boyle's law.

Question 3

For an adiabatic process, the relation between temperature and volume is:

Accepted Answer

TV^{\gamma - 1} = 	ext{constant}. Substituting P = \dfrac{nRT}{V} into PV^\gamma =  constant gives TV^{\gamma-1} =  constant.

Question 4

For an ideal gas undergoing an adiabatic process, the bulk modulus of elasticity is equal to:

Accepted Answer

\gamma P. Differentiating PV^\gamma =  constant gives \dfrac{dP}{dV} = -\gamma\dfrac{P}{V}, so K = -V\dfrac{dP}{dV} = \gamma P.

Question 5

The isothermal bulk modulus of elasticity of an ideal gas at pressure P is:

Accepted Answer

P. Differentiating PV =  constant gives \dfrac{dP}{dV} = -\dfrac{P}{V}, so K_{iso} = -V\dfrac{dP}{dV} = P.

Question 6

The work done by n moles of an ideal gas during an isothermal expansion from volume V_1 to V_2 at temperature T is:

Accepted Answer

nRT\ln\dfrac{V_2}{V_1}. W = \int_{V_1}^{V_2} P\,dV = \int_{V_1}^{V_2}\dfrac{nRT}{V}dV = nRT\ln\dfrac{V_2}{V_1}.

Question 7

The work done by an ideal gas during an adiabatic change from state (P_1, V_1, T_1) to (P_2, V_2, T_2) is:

Accepted Answer

\dfrac{P_1V_1 - P_2V_2}{\gamma - 1}. With Q = 0, W = -\Delta U = nC_v(T_1 - T_2) = \dfrac{nR(T_1 - T_2)}{\gamma - 1} = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1}.

Question 8

An ideal gas expands from V_0 to 2V_0, once isothermally doing work W_1 and once adiabatically doing work W_2. Then:

Accepted Answer

W_1 > W_2. Starting from the same state, the adiabatic curve falls more steeply than the isothermal, so it encloses less area under it. Hence the isothermal expansion does more work.

Question 9

When an ideal gas expands adiabatically, its temperature:

Accepted Answer

Falls, because the gas does work at the cost of its internal energy. With Q = 0 the work of expansion can only be paid for out of internal energy, so U and therefore T decrease.

Question 10

A monatomic ideal gas (\gamma = 5/3) expands adiabatically so that its volume increases by a factor of 8. Its absolute temperature decreases by a factor of:

Accepted Answer

4. TV^{\gamma-1} =  constant gives T_2 = T_1\left(\dfrac{1}{8}\right)^{2/3} = T_1\left(\dfrac{1}{2}\right)^2 = \dfrac{T_1}{4}.

Question 11

On a P-V diagram drawn from the same initial state, the adiabatic curve is:

Accepted Answer

Steeper than the isothermal curve. The slopes are -\dfrac{\gamma P}{V} for the adiabatic and -\dfrac{P}{V} for the isothermal. Since \gamma > 1, the adiabatic is steeper.

Question 12

For a process to be very nearly isothermal, it must be carried out:

Accepted Answer

Slowly, in a vessel with perfectly conducting walls. Heat must have time to flow in or out to hold the temperature fixed, which requires both conducting walls and a slow change.

Question 13

For a process to be very nearly adiabatic, it must be carried out:

Accepted Answer

Rapidly, in a vessel with perfectly insulating walls. Heat exchange must be prevented, which is achieved by insulating the vessel and completing the change so fast that there is no time for heat to flow.

Question 14

The sudden bursting of a cycle tyre is an example of a process that is very nearly:

Accepted Answer

Adiabatic. The expansion is far too rapid for any appreciable heat exchange, so Q \approx 0 and the escaping air cools noticeably.

Question 15

A gas with \gamma = 1.5 expands adiabatically until its volume is doubled. The ratio of its final to initial absolute temperature is:

Accepted Answer

0.707. \dfrac{T_2}{T_1} = \left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = \left(\dfrac{1}{2}\right)^{0.5} = \dfrac{1}{\sqrt{2}} \approx 0.707.

Question 16

For an adiabatic process the relation between pressure and temperature is:

Accepted Answer

P^{1-\gamma}T^\gamma = 	ext{constant}. Eliminating V between PV^\gamma =  constant and PV = nRT gives P^{1-\gamma}T^\gamma =  constant.

Question 17

During an isothermal compression of an ideal gas, heat is:

Accepted Answer

Rejected by the gas to the surroundings. With \Delta U = 0 and W negative (work done on the gas), Q = W is also negative, so the gas gives out heat to stay at constant temperature.

Question 18

One mole of an ideal gas expands isothermally at 300	ext{ K} until its volume doubles. The work done is (R = 8.314, \ln 2 = 0.693):

Accepted Answer

1729	ext{ J}. W = nRT\ln\dfrac{V_2}{V_1} = 1 	imes 8.314 	imes 300 	imes 0.693 \approx 1729	ext{ J}.

Question 19

A gas with \gamma = \dfrac{4}{3} expands adiabatically so that its volume becomes 27 times the original. Its absolute temperature becomes:

Accepted Answer

\dfrac{1}{3} of the original. TV^{\gamma-1} =  constant with \gamma - 1 = \dfrac{1}{3}, so T_2 = T_1\left(\dfrac{1}{27}\right)^{1/3} = \dfrac{T_1}{3}.

Question 20

A diatomic gas (\gamma = 1.4) at pressure P_1 expands adiabatically until its volume doubles. Its final pressure is approximately:

Accepted Answer

0.38P_1. P_2 = P_1\left(\dfrac{V_1}{V_2}\right)^\gamma = P_1(0.5)^{1.4} \approx 0.379P_1.

Isothermal & Adiabatic Processes — MH-CET Physics MCQs with Solutions

Questions with solutions