JEE MainPhysicsModerateThermodynamics

Select the correct alternative for an ideal gas

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\Delta U = n\frac{f}{2}R\left( T_{2} - T_{1} \right) = nC_{v}\left( T_{2} - T_{1} \right)

For adiabatic process

Q_{12} = 0

Q_{12} = \Delta U + W_{12}

W_{12} = - \Delta U

For Isothermal process Temperature is constant.
So, change internal energy is zero.

Source: Grass-G / P-08 Physics paper 23 October.docx

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