A sample of an ideal gas undergoes an isother | vedclass

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(B) For an isothermal process,the temperature $T$ remains constant $(dT = 0)$.Since the internal energy $U$ of an ideal gas depends only on temperatur

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$dQ = +ve, dU = 0, dW = +ve$

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(B) For an isothermal process,the temperature $T$ remains constant $(dT = 0)$.Since the internal energy $U$ of an ideal gas depends only on temperature $(U = f(T))$,the change in internal energy $dU = 0$.During expansion,the gas does work on the surroundings,so the work done by the gas is positive $(dW > 0)$.According to the first law of thermodynamics,$dQ = dU + dW$.Substituting the values,$dQ = 0 + dW$,which implies $dQ = dW$.Since $dW > 0$,it follows that $dQ > 0$ (heat is supplied to the system).Therefore,$dQ = +ve$,$dU = 0$,and $dW = +ve$.

$A$ sample of an ideal gas undergoes an isothermal expansion. If $dQ$,$dU$,and $dW$ represent the amount of heat supplied,the change in internal energy,and the work done respectively,then:

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