JEE PHYSICS MCQ: THERMODYNAMICS

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Application +4 | -1

A certain mass of gas at

273K

is expanded to 81 times its volume under adiabatic condition. If

\gamma = 1.25

for the gas, then its final temperature is

View answer and explanation

Correct option: B

- {182^ \circ }C

Option B (

- {182^ \circ }C

) is correct because it is calculated using the adiabatic process equation

T{V^{\gamma - 1}} = constant

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0.25\,\,{m^3}

{10^3}N/{m^2},

T

P

V

\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)