JEE PHYSICS MCQ: THERMODYNAMICS

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Application +4 | -1

n

moles of an ideal gas undergo a process in which the temperature changes with volume as

T=K V^2

. The work done by the gas as the temperature changes from

T_0

4 T_0

View answer and explanation

Correct option: C

\left(\frac{3}{2}\right) n R T_0

Option C (

\left(\frac{3}{2}\right) n R T_0

) is correct because the work done is calculated by integrating p dV, using the given temperature-volume relationship.

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0.25\,\,{m^3}

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

T

P

V

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