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].

_{0}. 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 P_{0}V_{0} ln 4

(C) Pressure at C is P_{0}/4

(D) Temperature at C is T_{0}/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(V_{2}/V_{1}) 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, V_{1 }is the initial volume and V_{2} is the final volume.

Therefore W = 1×RT_{0} ln (V_{2}/V_{1}) = RT_{0} ln(4V_{0}/V_{0})

Since RT_{0} = P_{0}V_{0} we obtain

W = P_{0}V_{0} 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 T_{0}/4. Considering the states at A and C we have

P_{0}V_{0 }/T_{0} = P_{C}V_{0}/(T_{0}/4)

The pressure at C is therefore given by

P_{C} = P_{0}/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_{i}V^{γ–1} = aT_{i}(V/32)^{γ–1}

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

T_{i}V^{2/5} = aT_{i}(V/32)^{2/5}

This gives a = 32^{2/5} = [2^{5}]^{2/5 }= 2^{2} = 4.

## No comments:

## Post a Comment