In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process $\mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{C} \rightarrow \mathrm{D}$ are $\mathrm{T}_{1}$ and $T_{2}$ ($T_{1}>T_{2}$) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

(B) $W_{\mathrm{AD}}=W_{\mathrm{BC}}$

(A) $\mathrm{W}_{\mathrm{AB}}=\mathrm{W}_{\mathrm{DC}}$

(C) $W_{\mathrm{AB}}<W_{\mathrm{CD}}$

(D) $W_{\mathrm{BC}}+W_{\mathrm{DA}}>0$

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Heat & Thermodynamics

In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process

A \to B

and

C \to D

are

T_{1}

and

T_{2}

(

T_{1} > T_{2}

) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

Answer: (B)

The gas is at

T_{1}

all along

A \to B

and at

T_{2}

all along

C \to D

, so

A

and

B

sit at

T_{1}

while

C

and

D

sit at

T_{2}

. For an adiabatic change the work done equals the fall in internal energy,

W = n C_{V} (T_{i} - T_{f})

, so

W_{A \to D} = n C_{V} (T_{1} - T_{2}) = W_{B \to C}

The same formula gives

W_{B C} + W_{D A} = n C_{V} (T_{1} - T_{2}) + n C_{V} (T_{2} - T_{1}) = 0

, so the two adiabatic works cancel rather than add to a positive number. On the isothermals

W_{A B} = n R T_{1} ln \frac{V _{B}}{V _{A}}

is positive while

W_{C D} = n R T_{2} ln \frac{V _{D}}{V _{C}}

is negative, and the ratio

W_{A B} / W_{D C} = T_{1} / T_{2} \neq = 1

. The relation that survives is

W_{A D} = W_{B C}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option