JEE PHYSICS MCQ: THERMODYNAMICS

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Application +4 | -1

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is

T{V^x} =

constant, then

x

View answer and explanation

Correct option: A

\frac{2}{5}

Option A (

\frac{2}{5}

) is correct because it is the value of x derived from the adiabatic process equation for a diatomic gas.

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