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(A) For an ideal gas,the change in internal energy $\Delta U$ is given by $\Delta U = \frac{f}{2} \mu R \Delta T$.Using the ideal gas equation $PV = \

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$10$

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(A) For an ideal gas,the change in internal energy $\Delta U$ is given by $\Delta U = \frac{f}{2} \mu R \Delta T$.Using the ideal gas equation $PV = \mu RT$,we can write $\Delta U = \frac{f}{2} (P_B V_B - P_A V_A)$.Here,the gas is diatomic,so the degrees of freedom $f = 5$.The number of moles $\mu = 1$.From the graph,at point $A$: $P_A = 8 	imes 10^3 \ Pa$ and $V_A = 2 \ m^3$.At point $B$: $P_B = 4 	imes 10^3 \ Pa$ and $V_B = 5 \ m^3$.Substituting these values:$\Delta U = \frac{5}{2} (P_B V_B - P_A V_A)$$\Delta U = \frac{5}{2} [(4 	imes 10^3 	imes 5) - (8 	imes 10^3 	imes 2)]$$\Delta U = \frac{5}{2} [20 	imes 10^3 - 16 	imes 10^3]$$\Delta U = \frac{5}{2} [4 	imes 10^3]$$\Delta U = 5 	imes 2 	imes 10^3 \ J = 10 	imes 10^3 \ J = 10 \ kJ$.

$A$ 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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