One mole of an ideal gas at 900 K undergoes | vedclass

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(D) From the graph,process $I$ is a vertical line,meaning it is an isentropic (reversible adiabatic) process.For process $I$ (adiabatic): $\Delta U_I

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$10$

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(D) From the graph,process $I$ is a vertical line,meaning it is an isentropic (reversible adiabatic) process.For process $I$ (adiabatic): $\Delta U_I = W_I$.Given $\frac{U}{R}$ changes from $2250 \ K$ to $450 \ K$,so $\Delta U_I = R(450 - 2250) = -1800 \ R$.Thus,$W_I = -1800 \ R$.For process $II$,it is a horizontal line,meaning it is an isothermal process (since $U$ depends only on $T$ for an ideal gas,constant $U$ implies constant $T$).For an isothermal process,$W_{II} = -nRT_2 \ln \frac{V_3}{V_2}$.Since $U = nC_{V, m}T$,at state $2$,$\frac{U_2}{R} = 450 = n 	imes \frac{5}{2} 	imes T_2$. With $n=1$,$T_2 = \frac{450 	imes 2}{5} = 180 \ K$.Given $W_I = W_{II}$,we have $-1800 \ R = -1 	imes R 	imes 180 \ln \frac{V_3}{V_2}$.$\ln \frac{V_3}{V_2} = \frac{1800}{180} = 10$.

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