Consider the P-V diagram for 1 mole of an ide | vedclass

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(A, B, C) The process is $A \rightarrow B \rightarrow C$. The coordinates are $A(2V_{0}, P_{0})$, $B(V_{0}, P_{0})$, and $C(V_{0}, 2P_{0})$.For an ide

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Change in internal energy for process $A \rightarrow B$ is $-\frac{3}{2} P_{0} V_{0}$

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(A, B, C) The process is $A \rightarrow B \rightarrow C$. The coordinates are $A(2V_{0}, P_{0})$, $B(V_{0}, P_{0})$, and $C(V_{0}, 2P_{0})$.For an ideal gas, $\Delta U = n C_{v} \Delta T = \frac{n C_{v}}{nR} (P_{f}V_{f} - P_{i}V_{i}) = \frac{C_{v}}{R} (P_{f}V_{f} - P_{i}V_{i})$.For a monatomic gas, $C_{v} = \frac{3}{2}R$, so $\Delta U = \frac{3}{2} (P_{f}V_{f} - P_{i}V_{i})$.For the whole process $A \rightarrow C$, $\Delta U = \frac{3}{2} (P_{C}V_{C} - P_{A}V_{A}) = \frac{3}{2} (2P_{0}V_{0} - P_{0}2V_{0}) = 0$. Thus, option $A$ is correct.Work done $W = \int P dV$. The area under the $P-V$ curve for $A \rightarrow B$ is $P_{0}(V_{0} - 2V_{0}) = -P_{0}V_{0}$. For $B \rightarrow C$, $dV = 0$, so $W = 0$. Total work $W = -P_{0}V_{0}$.From the first law, $\Delta Q = \Delta U + W$. Since $\Delta U = 0$ and $W For process $A \rightarrow B$, $\Delta U_{AB} = \frac{3}{2} (P_{B}V_{B} - P_{A}V_{A}) = \frac{3}{2} (P_{0}V_{0} - P_{0}2V_{0}) = -\frac{3}{2} P_{0}V_{0}$. Thus, option $C$ is correct.As calculated, total work is $-P_{0}V_{0}$, so option $D$ is incorrect.

Consider the $P-V$ diagram for $1$ mole of an ideal monatomic gas shown in the figure. Which of the following statements is/are true?

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