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(B) For a monatomic ideal gas,the number of moles $n = 1$ and the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.From the $p-V$ diagr

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$\frac{13 R}{6}$

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(B) For a monatomic ideal gas,the number of moles $n = 1$ and the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.From the $p-V$ diagram,the coordinates are $A(V_0, 3p_0)$ and $B(5V_0, 6p_0)$.Using the ideal gas equation $pV = nRT$,the temperatures at $A$ and $B$ are:$T_A = \frac{p_A V_A}{nR} = \frac{(3p_0)(V_0)}{1 \cdot R} = \frac{3p_0 V_0}{R}$$T_B = \frac{p_B V_B}{nR} = \frac{(6p_0)(5V_0)}{1 \cdot R} = \frac{30p_0 V_0}{R}$The change in temperature is $\Delta T = T_B - T_A = \frac{30p_0 V_0}{R} - \frac{3p_0 V_0}{R} = \frac{27p_0 V_0}{R}$.The change in internal energy is $\Delta U = n C_v \Delta T = 1 \cdot \left(\frac{3}{2} R\right) \cdot \left(\frac{27p_0 V_0}{R}\right) = \frac{81}{2} p_0 V_0$.The work done $W$ is the area under the $p-V$ graph,which is a trapezoid:$W = \frac{1}{2} (p_A + p_B) (V_B - V_A) = \frac{1}{2} (3p_0 + 6p_0) (5V_0 - V_0) = \frac{1}{2} (9p_0) (4V_0) = 18p_0 V_0$.Using the first law of thermodynamics,$Q = \Delta U + W$:$Q = \frac{81}{2} p_0 V_0 + 18 p_0 V_0 = \frac{81 + 36}{2} p_0 V_0 = \frac{117}{2} p_0 V_0$.Since $Q = n C \Delta T$,where $n = 1$:$C = \frac{Q}{\Delta T} = \frac{117/2 \cdot p_0 V_0}{27 p_0 V_0 / R} = \frac{117}{2} \cdot \frac{R}{27} = \frac{117}{54} R = \frac{13}{6} R$.

One mole of a monatomic ideal gas undergoes the process $A \rightarrow B$ in the given $p-V$ diagram. The specific heat capacity in the process is:

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