JEE PHYSICS MCQ: THERMODYNAMICS

1 Easy

1 of 1

Application +4 | -1

A gas expand

0.25\,\,{m^3}

at constant pressure

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

the work done is

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Correct option: B

250 J

Option B (250 J) is correct because it accurately calculates the work done during the gas expansion at constant pressure.

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

\left( {\gamma = 5/3} \right)