Assertion : For an isothermal reversible proc | vedclass

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(B) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas depends only on temperature,$\Delta

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If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

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(B) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas depends only on temperature,$\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q + W$.Substituting $\Delta U = 0$,we get $0 = Q + W$,which implies $Q = -W$.This means the heat absorbed by the system is equal to the work done by the system.However,for an isothermal process,enthalpy change $\Delta H = \Delta U + \Delta(PV) = 0 + \Delta(nRT)$. Since $n, R, T$ are constant,$\Delta H = 0$ for an ideal gas.While $\Delta H = 0$ is true for an ideal gas in an isothermal process,the reason for $Q = -W$ is the first law of thermodynamics $(\Delta U = 0)$,not the fact that $\Delta H = 0$.

Assertion : For an isothermal reversible process $Q = -W$,i.e.,work done by the system equals the heat absorbed by the system.
Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.

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Assertion : For an isothermal reversible process $Q = -W$,i.e.,work done by the system equals the heat absorbed by the system.Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.