IIT-JEE Multiple Correct Answer Type Questions on Thermodynamics

"Hate the sin, love the sinner."
–Mahatma Gandhi

Today we will discuss two multiple correct answer type questions on thermodynamics. These were included in the IIT-JEE 2009 question paper:

(1) The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

(A) the process during the path A → B is isothermal

(B) heat flows out of the gas during the path B → C → D

(C) work done during the path A → B → C is zero

(D) positive work is done by the gas in the cycle ABCDA.

Statement (A) is incorrect since a curve representing an isothermal in a PV diagram must be a hyperbola and not an arc of a circle.

We have PV = nRT with usual notations. The value of V is smaller at D compared to that at B where as the value of P is the same. The product PV = nRT is therefore smaller at D. This means that the temperature at D is less than that at B and hence statement (B) is correct.

During the path AB the gas expands, doing work. During the path BC the gas is compressed and work is done on the gas. But these works are unequal since the area under AB is greater than the area under BC. Therefore, statement (c) is incorrect.

The cycle of operations ABCDA is clockwise and the cyclic curve encloses an area. Therefore, positive work is done by the gas and statement (D) is correct.

So the correct options are (B) and (D).

(2) C_V and C_P denote the molar specific heat capacities of a gas at constant volume and constant pressure respectively. Then

(A) C_P – C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(B) C_P + C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(C) C_P/C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(D) C_P.C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas.

In the case of ideal gases C_P – C_V = R, the universal gas constant. Therefore statement (A) is wrong.

C_P and C_V are larger for a diatomic ideal gas compared to a mono-atonic ideal gas. Therefore, C_P + C_V is larger for a diatomic ideal gas. Statement (B) is correct.

[For diatomic ideal gas C_P = 7R/2 and C_V = 5R/2. For mono-atomic ideal gas C_P = 5R/2 and C_V = 3R/2].

Obviously statement (C) is wrong.

Statement (D) is correct since both C_P and C_V are larger for a diatomic ideal gas than for a mono-atomic ideal gas.

Therefore, options (B) and (D) are correct.

You will find a useful post on thermodynamics here.

IIT-JEE 2010 Questions on Thermodynamics (Multiple Correct Answer Type and Integer Type)

The following questions on thermodynamics appeared in the IIT-JEE 2010 question paper:

[Question No.1 is Multiple Correct Answer Type. Question No.2 is Integer Type, the answer to which is a single digit integer ranging from 0 to 9].

(1) One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, As shown in figure. Its pressure at A is P₀. Choose the correct option(s) from the following:

(A) Internal energies at A and B are the same

(B) Work done by the gas in process AB is P₀V₀ ln 4

(C) Pressure at C is P₀/4

(D) Temperature at C is T₀/4

AB is an isothermal process (A and B are at the same temperature). Therfore the internal energies at A and B are the same.

The work (W) done by the gas in the isothermal process AB is given by

W = nRT ln(V₂/V₁) where R is universal gas constant, ‘n’ is the number of moles in the sample of the gas, T is the temperature at which the process occurs, V₁ is the initial volume and V₂ is the final volume.

Therefore W = 1×RT₀ ln (V₂/V₁) = RT₀ ln(4V₀/V₀)

Since RT₀ = P₀V₀ we obtain

W = P₀V₀ ln 4

Thus options A and B are correct.

We have PV/T = constant

Assuming that the line BC passes through the origin, the temperature at C must be T₀/4. Considering the states at A and C we have

P₀V₀ /T₀ = P_CV₀/(T₀/4)

The pressure at C is therefore given by

P_C = P₀/4

So all options are correct.

(2) A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is:

In the case of an adiabatic process we have

TV^γ–1 = constant where γ is the ratio of specific heats of the gas.

Therefore, T_iV^γ–1 = aT_i(V/32)^γ–1

For an ideal diatomic gas γ = 7/5 and hence we have

T_iV^2/5 = aT_i(V/32)^2/5

This gives a = 32^2/5 = [2⁵]^2/5 = 2² = 4.

Questions on Isothermal and Adiabatic Changes

You can expect questions involving isothermal and adiabatic processes in most entrance examinations. Here is a typical question:
A gas at a temperature of 27°C inside a container is suddenly compressed to one sixteenths of its initial volume. The temperature of the gas immediately after the compression is (Ratio of specific heats of the gas, γ = 1.5)
(a) 19200 K (b) 1200°C (c) 927°C (d) 108°C (e) 19200°C
This is an adiabatic change since the compression is sudden so that the volume(V) and the temperature (T) are related as TV^γ–1 = constant.
Therefore we have 300 V^0.5 = T(V/16)^0.5. Note that the temperature is to be substituted in Kelvin. The final temperature is given by T = 300×16^0.5 = 1200 K = 927°C.
The following MCQ is meant for testing your understanding of the work done in thermodynamic processes:

Starting from the same initial conditions an ideal gas expands from volume V₁ to volume V₂ in three different ways: (i) Adiabatically, doing work W₁. (ii) Isothermally, doing work W₂. (iii) Isobarically, doing work W₃. Then,

(a) W₁ = W₂ = W₃ (b) W₁ > W₂ > W₃ (c) W₃ > W₂ > W₁ (d) W₂ > W₁ > W₃ (e) W₁ > W₃ > W₂

Since the work done is the area under the corresponding curve in a PV diagram, you can easily verify that the work done is the largest in the isobaric case since it is a straight line parallel to the volume axis. The adiabatic curve is the steepest one so that the area under it is the smallest. The correct option therefore is (c).
You can easily show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve as follows:
In the case of an adiabatic change, the pressure and volume are related as PV^γ = constant.
Differentiating, P γV^γ–1 dV + V^γdP = 0
Slope of adiabatic curve = dP/dV = (– γPV^γ–1)/V^γ = – γP/V
In the case of an isothermal change, the pressure and volume are related as PV= constant.
Differentiating, PdV + VdP = 0.
Slope of isothermal curve = dP/dV = – P/V.
This show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve.
The following MCQ also pertains to adiabatic compression:
A gas having a volume of 800 cm³ is suddenly compressed to 100cm³. If the initial pressure is P, the final pressure is (γ = 5/3)
(a) P/32 (b) 24P (c) 32P (d) 8P (e) 16P
We have PV^γ = constant so that P×800^5/3 = P'×100^5/3 from which P' = P×8^5/3 = P×2⁵ = 32P.

Now, see this simple question:
An ideal gas expands isothermally from volume V₁ to volume V₂. It is then compressed to the original volume V₁ adiabatically. The initial pressure is P₁, final pressure is P₂ and the net work done by the gas during the entire process is W. Then
(a) P₁ = P₂, W>0 (b) P₁> P₂, W>0 (c) P₂ > P₁, W>0 (d) P₂ > P₁, W=0 (e) P₂ > P₁, W<0

The adiabatic compression will increase the temperature of the gas so that the final pressure (P₂) when the volume is restored to the value V₁ is greater than the initial pressure P₁. Since the pressure is greater during the adiabatic compression, more work has to be done on the gas. The work done on the gas is thus greater than the work done by the gas. In other words, the net work done by the gas during the entire process is negative. So, the correct option is (e).