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Question

An ideal monatomic gas is taken from state A to B, then from state B to state C and finally back to state A as shown in the p-V diagram. Which of the following is correct for BC process?

Moderate

A

ΔQ +ve, ΔU -ve, ΔW +ve
B

ΔQ +ve, ΔU+-ve, ΔW -ve
C

ΔQ +ve, ΔU +ve, ΔW +ve
D

ΔQ -ve, ΔU -ve, ΔW -ve

Solution

T₀ = T_A = p₀V₀/nR, T_B = 5T₀, T_C = 4T₀
So for BC, ΔU_BC = nC_V(T_C-T_B)
ΔU_BC = n 3R/2 (4T₀-5T₀)
$= n\frac{3}{2}R\left( { - \frac{{{p_0}{v_0}}}{{nR}}} \right) = - \frac{3}{2}{p_0}{v_0}$
Also $\Delta {W_{BC}} = - \frac{1}{2}\left( {4{p_0} + {p_0}} \right)\left( {5{v_0} - {v_0}} \right) = - 10{p_0}{v_0}$
$\therefore \Delta {Q_{BC}} = \Delta {U_{BC}} + \Delta {W_{BC}} = - 11.5{p_0}{v_0}$

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