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(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = \frac{f}{2} n R \Delta T = \frac{f}{2} (P_2 V_2 - P_1 V_1)$.For

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$-20 	ext{ kJ}$

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(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = \frac{f}{2} n R \Delta T = \frac{f}{2} (P_2 V_2 - P_1 V_1)$.For a diatomic gas,the degrees of freedom $f = 5$.From the given $P-V$ diagram,at point $A$,$P_1 = 5 	ext{ kPa} = 5 	imes 10^3 	ext{ Pa}$ and $V_1 = 4 	ext{ m}^3$.At point $B$,$P_2 = 2 	ext{ kPa} = 2 	imes 10^3 	ext{ Pa}$ and $V_2 = 6 	ext{ m}^3$.Substituting these values into the formula:$\Delta U = \frac{5}{2} (P_2 V_2 - P_1 V_1)$$\Delta U = \frac{5}{2} [(2 	imes 10^3 	imes 6) - (5 	imes 10^3 	imes 4)]$$\Delta U = \frac{5}{2} [12 	imes 10^3 - 20 	imes 10^3]$$\Delta U = \frac{5}{2} [-8 	imes 10^3]$$\Delta U = 5 	imes (-4 	imes 10^3) = -20 	imes 10^3 	ext{ J} = -20 	ext{ 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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