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(A) In the given process,for an ideal gas,$\Delta W = 0$ and $\Delta Q < 0$.According to the first law of thermodynamics,$\Delta Q = \Delta W + \Delta

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the temperature will decrease

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(A) In the given process,for an ideal gas,$\Delta W = 0$ and $\Delta Q According to the first law of thermodynamics,$\Delta Q = \Delta W + \Delta U$,where $\Delta U$ is the change in internal energy of the gas.Since $\Delta W = 0$,we have $\Delta Q = \Delta U$.Given that $\Delta Q For an ideal gas,the internal energy $U$ is a function of temperature $T$ only $(U = nC_vT)$.Therefore,a decrease in internal energy $(\Delta U < 0)$ implies a decrease in the temperature of the gas.

In a given process,for an ideal gas,$\Delta W = 0$ and $\Delta Q < 0$. Then,for the gas,

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