Physics · Thermodynamics

JEE Advanced 2020 — Paper 1 — Question 15

Consider one mole of helium gas enclosed in a container at initial pressure $P_{1}$ and volume $V_{1}$ . It expands isothermally to volume $4 V_{1}$ . After this, the gas expands adiabatically and its volume becomes $32 V_{1}$ . The work done by the gas during isothermal and adiabatic expansion processes are $W_{\text {iso }}$ and $W_{\text {adia }}$ , respectively. If the ratio $\frac{W_{\text {iso }}}{W_{\text {adia }}}=f \ln 2$ , then $f$ is $\_\_\_\_$ .

Answer: 1.77

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{A} \rightarrow \mathrm{B}$

$\mathrm{P}_{1} \mathrm{~V}_{1}=\mathrm{P}_{\mathrm{B}} \times 4 \mathrm{~V}_{1}$

$\Rightarrow P_{B}=\frac{P_{1}}{4}$

$W_{\text {iso }}=1 \times R \times T \ell n \frac{4 V_{1}}{V_{1}}$

$P_{1} V_{1} \ell n 4=2 P_{1} V_{1} \ell n 2$

$\mathrm{B} \rightarrow \mathrm{C}$ $P_{B} V_{B}^{\gamma}=P_{C} V_{C}^{\gamma}$ $\frac{P_{1}}{4}\left(4 \mathrm{~V}_{1}\right)^{\gamma}=P_{C}\left(32 \mathrm{~V}_{1}\right)^{\gamma}$

$\Rightarrow P_{C}=\frac{1}{4}\left(\frac{1}{8}\right)^{\gamma} P_{1}=\frac{1}{4}\left(\frac{1}{2^{3}}\right)^{5 / 3} P_{1}=\frac{P_{1}}{128}$

$W_{\text {adic }}=\frac{\frac{P_{1}}{128} \times 32 \mathrm{~V}_{1}-\frac{P_{1}}{4} \times 4 \mathrm{~V}_{1}}{1-\frac{5}{3}}$

$=\frac{\frac{P_{1} V_{1}}{4}-P_{1} V_{1}}{-\frac{2}{3}}=\frac{-\frac{3}{4} P_{1} V_{1}}{-\frac{2}{3}}=\frac{9}{8} P_{1} V_{1}$

$\frac{W_{\text {iso }}}{W_{\text {adio }}}=\frac{2 P_{1} V_{1} \ell n 2}{\frac{9}{8} P_{1} V_{1}}=\frac{16}{9} \ell n 2$

Comparing with given expression of $=16 / 9=1.78$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Physics
Chapter: Thermodynamics
Topic: Efficiency of Processes and Miscellaneous Problems