Assertion : For an isothermal reversible proc | vedclass

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(C) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas is a function of temperature only,$

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If the Assertion is correct but Reason is incorrect.

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(C) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas is a function of temperature only,$\Delta U = 0$.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 $\Delta H = \Delta U + \Delta(PV)$. For an ideal gas,$\Delta H = \Delta U + \Delta(nRT) = 0 + nR\Delta T = 0$. While $\Delta H$ is zero for an ideal gas,the reason provided is not the fundamental explanation for $Q = -W$,which is based on the first law and $\Delta U = 0$. In general,$\Delta H$ is not necessarily zero for all isothermal processes (e.g.,real gases). Thus,the Assertion is correct,but the Reason is incorrect.

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.