The graph shows a P-V diagram where A-B is an | vedclass

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(C) In a $P-V$ diagram, the internal energy $U$ of an ideal gas depends only on temperature $T$ $(U = nC_vT)$. The isotherms are curves of constant te

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Process $AC$ is endothermic, $BC$ is exothermic.

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(C) In a $P-V$ diagram, the internal energy $U$ of an ideal gas depends only on temperature $T$ $(U = nC_vT)$. The isotherms are curves of constant temperature. For an ideal gas, $PV = nRT$, so $T = PV/nR$. $1$. Consider the adiabatic curve $AB$. Since it is an adiabatic expansion, the temperature at $A$ $(T_A)$ is greater than the temperature at $B$ $(T_B)$. $2$. For process $AC$: The path moves from $A$ to $C$. Since $C$ lies above the adiabatic curve $AB$, the temperature at $C$ $(T_C)$ is greater than the temperature at any point on the adiabatic curve $AB$ at the same volume. Specifically, $T_C > T_A$. Since the temperature increases, the internal energy increases $(\Delta U > 0)$. Since the gas expands from $A$ to $C$, work is done by the gas $(W > 0)$. By the first law of thermodynamics, $Q = \Delta U + W$. Since both $\Delta U$ and $W$ are positive, $Q > 0$, meaning the process is endothermic. $3$. For process $BC$: The path moves from $B$ to $C$. Since $C$ is at a higher temperature than $B$ $(T_C > T_B)$, the internal energy increases $(\Delta U > 0)$. However, the path $BC$ is a compression (volume decreases), so work is done on the gas $(W < 0)$. In this specific geometry, the heat $Q$ is negative because the energy required to increase the temperature is less than the energy released by compression, or more simply, $C$ lies below the isotherm passing through $B$. Thus, $BC$ is exothermic.

The graph shows a $P-V$ diagram where $A-B$ is an adiabatic curve. Choose the correct statement regarding the processes $AC$ and $BC$.

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