Starting at temperature 300 K, one mole of an ideal diatomic gas ( γ… | JEE Main 2020 (9 Jan, Shift 2)

Q210JEE Main 2020 · 9 JAN · SHIFT 2Difficult

Thermodynamic Processes

Starting at temperature 300 K, one mole of an

ideal diatomic gas ( $\gamma$ = 1.4) is first compressed

adiabatically from volume V1 to V2 = $\frac{V_1}{16}$ . It is

then allowed to expand isobarically to volume 2V2. If all the processes are the quasi-static then the final temperature of the gas (in oK) is (to the nearest integer) _____.

Numerical answer type. Work it out, then open the answer.

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Answer: 1818

Adiabatic: $T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \times 16^{0.4}$
$16^{0.4} = 2^{1.6} \approx 3.03$ , so $T_2 \approx 909.4\text{ K}$ .
Isobaric: $V \propto T$ , so doubling volume doubles temperature: $T_3 = 2 \times 909.4 \approx 1819\text{ K}$ .
Final temperature $\approx 1818$ – $1819$ K.

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