JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Heat & Thermodynamics

Starting at temperature 300 K, one mole of an ideal diatomic gas (

γ = 1.4

) is first compressed adiabatically from volume

V_{1}

V_{2} = \frac{V _{1}}{16}

. It is then allowed to expand isobarically to volume

2 V_{2}

. If all the processes are the quasi-static then the final temperature of the gas (in

^{\circ}

K) is (to the nearest integer) ____.

Answer: 1819

Solution

The adiabatic compression obeys

T V^{γ - 1} = constant

, so

T_{B} = 300 (\frac{V _{1}}{V _{2}})^{0.4} = 300 \times 1 6^{0.4} = 300 \times 2^{1.6}

. The gas is then expanded at constant pressure from

V_{2}

2 V_{2}

, where

\frac{V}{T}

is constant, so the temperature doubles:

T_{C} = 2 T_{B} = 600 \times 2^{1.6}

. Since

2^{1.6} \approx 3.03

, the final temperature is

T_{C} \approx 600 \times 3.03 \approx 1819 K

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Integer answer