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JEE MAIN 2020
09-01-2020 S2
Question
Starting at temperature 300 K , one mole of an ideal diatomic gas ( $\gamma=1.4$ ) is first compressed adiabatically from volume $V_1$ to $V_2=\frac{V_1}{16}$. It is then allowed to expand isobarically to volume $2 \mathrm{~V}_2$. If all the processes are the quasi-static then the final temperature of the gas (in ${ }^{\circ} \mathrm{K}$ ) is (to the nearest integer) $\_\_\_\_$ .
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Solution
$\begin{aligned} & T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \\ & T_2=300 \times(16)^{0.4} \\ & T_f=2 T_2=300 \times 2 \times(16) 0.4 \\ & =1818 \mathrm{~K}\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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