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(B) We know that the change in internal energy for an ideal gas is given by $\Delta U = n C_v \Delta T$.For a diatomic gas,the molar heat capacity at

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$-20$

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(B) We know that the change in internal energy 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 $C_v = \frac{5R}{2}$.Substituting this into the equation,we get $\Delta U = n \left( \frac{5R}{2} ight) (T_B - T_A)$.Using the ideal gas law $PV = nRT$,we can write $T = \frac{PV}{nR}$,so $\Delta U = \frac{5}{2} (P_B V_B - P_A V_A)$.From the given graph,at point $A$: $P_A = 5 	imes 10^3 \, Pa$ and $V_A = 4 \, m^3$.At point $B$: $P_B = 2 	imes 10^3 \, Pa$ and $V_B = 6 \, m^3$.Substituting these values:$\Delta U = \frac{5}{2} [(2 	imes 10^3 	imes 6) - (5 	imes 10^3 	imes 4)]$$\Delta U = \frac{5}{2} [12 	imes 10^3 - 20 	imes 10^3]$$\Delta U = \frac{5}{2} [-8 	imes 10^3]$$\Delta U = -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 ............ $kJ$.

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