Given below are two statements:Statement-I: F | vedclass

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(A) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = q

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Both Statement-$I$ and Statement-$II$ are correct

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(A) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = q + w$.Substituting $\Delta U = 0$,we get $q = -w$.For an irreversible process,$w = -P_{	ext{ext}} \Delta V = -P_{	ext{ext}} (V_{	ext{final}} - V_{	ext{initial}})$.Thus,$q = -w = P_{	ext{ext}} (V_{	ext{final}} - V_{	ext{initial}})$. Statement-$I$ is correct.For an adiabatic process,the heat exchange $q = 0$.From the first law,$\Delta U = q + w = 0 + w_{	ext{adiabatic}} = w_{	ext{adiabatic}}$. Statement-$II$ is correct.

Given below are two statements:
Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$
Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$
The correct answer is

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Given below are two statements:Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$The correct answer is