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(C) The given adiabatic process follows $P^3 V^5 = 	ext{constant}$, which can be written as $P V^{5/3} = 	ext{constant}$. Comparing this with $P V^{

Vedclass · Accepted Answer

$588$

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(C) The given adiabatic process follows $P^3 V^5 = 	ext{constant}$, which can be written as $P V^{5/3} = 	ext{constant}$. Comparing this with $P V^{\gamma} = 	ext{constant}$, we get $\gamma = 5/3$.For an adiabatic process, $\Delta Q = 0$, so $\Delta U = -W$. The work done in the adiabatic process is:$W_a = \frac{P_f V_f - P_i V_i}{1 - \gamma} = \frac{(\frac{1}{32} 	imes 10^5 	imes 8 	imes 10^{-3}) - (10^5 	imes 10^{-3})}{1 - 5/3} = \frac{250 - 100}{-2/3} = \frac{150}{-2/3} = -225 	ext{ J}$.Thus, $\Delta U = -W_a = 225 	ext{ J}$.In the two-step process:$1$. Isobaric expansion from $(P_i, V_i)$ to $(P_i, V_f)$: $W_1 = P_i(V_f - V_i) = 10^5(8 	imes 10^{-3} - 10^{-3}) = 700 	ext{ J}$.$Q_1 = nC_p\Delta T = \frac{\gamma}{\gamma - 1} P_i(V_f - V_i) = \frac{5/3}{2/3} 	imes 700 = 2.5 	imes 700 = 1750 	ext{ J}$.$2$. Isochoric process from $(P_i, V_f)$ to $(P_f, V_f)$:$W_2 = 0$.$Q_2 = nC_v\Delta T = \frac{1}{\gamma - 1} V_f(P_f - P_i) = \frac{1}{2/3} 	imes 8 	imes 10^{-3} 	imes (\frac{1}{32} 	imes 10^5 - 10^5) = 1.5 	imes 8 	imes 10^{-3} 	imes (-31/32 	imes 10^5) = 1.5 	imes (-31/4) 	imes 100 = -1162.5 	ext{ J}$.Total heat $Q = Q_1 + Q_2 = 1750 - 1162.5 = 587.5 	ext{ J} \approx 588 	ext{ J}$.

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