JEE PHYSICS MCQ: THERMODYNAMICS

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Understanding +4 | -1

When a small amount of heat

\Delta Q

is added to an enclosed gas, then increase in internal energy and external work done are related as

View answer and explanation

Correct option: B

\Delta Q = m{C_v}\Delta T + p\Delta V

Option B (

\Delta Q = m{C_v}\Delta T + p\Delta V

) is correct because it represents the first law of thermodynamics, where heat added equals the change in internal energy plus the work done.

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