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(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_V \Delta T$.For a diatomic gas,the molar heat capacity at co

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$-20 \, kJ$

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(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_V \Delta T$.For a diatomic gas,the molar heat capacity at constant volume is $C_V = \frac{R}{\gamma - 1}$,where $\gamma = 1.4 = 7/5$.Using the ideal gas law $PV = nRT$,we have $\Delta T = \frac{P_B V_B - P_A V_A}{nR}$.Substituting this into the internal energy formula:$\Delta U = n \left( \frac{R}{\gamma - 1} ight) \left( \frac{P_B V_B - P_A V_A}{nR} ight) = \frac{P_B V_B - P_A V_A}{\gamma - 1}$.From the graph,at point $A$: $P_A = 5 	imes 10^3 \, Pa$,$V_A = 4 \, m^3$.At point $B$: $P_B = 2 	imes 10^3 \, Pa$,$V_B = 6 \, m^3$.Calculating the product $PV$ at each point:$P_A V_A = (5 	imes 10^3) 	imes 4 = 20 	imes 10^3 \, J$.$P_B V_B = (2 	imes 10^3) 	imes 6 = 12 	imes 10^3 \, J$.Now,$\Delta U = \frac{12 	imes 10^3 - 20 	imes 10^3}{7/5 - 1} = \frac{-8 	imes 10^3}{2/5} = -8 	imes 10^3 	imes \frac{5}{2} = -20 	imes 10^3 \, J = -20 \, kJ$.

One mole of an ideal diatomic gas undergoes a transition from $A$ to $B$ along a path $AB$ as shown in the figure. The change in internal energy of the gas during the transition is:

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