During an isothermal expansion of an ideal ga | vedclass

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(D) For an ideal gas,the internal energy $U$ is a function of temperature $T$ only,i.e.,$U = f(T)$.During an isothermal process,the temperature $T$ re

Vedclass · Accepted Answer

Both $(b)$ and $(c)$ are correct

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(D) For an ideal gas,the internal energy $U$ is a function of temperature $T$ only,i.e.,$U = f(T)$.During an isothermal process,the temperature $T$ remains constant.Therefore,the change in internal energy $\Delta U = 0$,which means the internal energy does not change.According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.Since $\Delta U = 0$,we have $\Delta Q = \Delta W$.This implies that the heat absorbed by the gas is equal to the work done by the gas.Thus,both statements $(b)$ and $(c)$ are correct.

During an isothermal expansion of an ideal gas,

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