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.