Q 11-11-134JEE MainJEE Main 2020 (9 Jan, Shift 2)Medium
Starting at temperature $300$ K, one mole of an ideal diatomic gas $(\gamma = 1.4)$ is first compressed adiabatically from volume $V_1$ to $V_2 = \dfrac{V_1}{16}$. It is then allowed to expand isobarically to volume $2V_2$. If all the processes are quasi-static then the final temperature of the gas (in K) is (to the nearest integer) ______.
Numerical value type. Enter your answer.
Answer: 1819
Adiabatic: $TV^{\gamma - 1} = $ constant.
$$T_2 = 300\times16^{0.4} = 300\times3.031 = 909.4\ \text{K}$$
Isobaric: $V \propto T$, so doubling the volume doubles the temperature:
$$T_3 = 2\times909.4 \approx 1819\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics