## Saturday, September 11, 2010

### 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 followin g:

(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 = PCV0/(T0/4)

The pressure at C is therefore given by

PC = 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

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

Therefore, TiVγ–1 = aTi(V/32)γ–1

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

TiV2/5 = aTi(V/32)2/5

This gives a = 322/5 = 2/5 = 22 = 4.