Two moles of an ideal monoatomic gas undergo | vedclass

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(B) Given: $n = 2 	ext{ moles}$,$T_3 = 2400 	ext{ K}$,$2 T_4 = 2400 \implies T_4 = 1200 	ext{ K}$,$3 T_2 = 2400 \implies T_2 = 800 	ext{ K}$,$6 T_

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$1600$

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(B) Given: $n = 2 	ext{ moles}$,$T_3 = 2400 	ext{ K}$,$2 T_4 = 2400 \implies T_4 = 1200 	ext{ K}$,$3 T_2 = 2400 \implies T_2 = 800 	ext{ K}$,$6 T_1 = 2400 \implies T_1 = 400 	ext{ K}$.From the graph,the process $1 	o 2$ is isochoric $(P = 	ext{constant})$,$2 	o 3$ is isobaric $(P \propto T)$,$3 	o 4$ is isochoric,and $4 	o 1$ is isobaric.Work done in a cycle $W = \oint P \, dV$. Using $PV = nRT$,$W = \oint nR \, dT$ for isobaric processes.$W_{12} = 0$ (isochoric).$W_{23} = nR(T_3 - T_2) = 2R(2400 - 800) = 3200R$.$W_{34} = 0$ (isochoric).$W_{41} = nR(T_1 - T_4) = 2R(400 - 1200) = -1600R$.Total work $W = W_{12} + W_{23} + W_{34} + W_{41} = 0 + 3200R + 0 - 1600R = 1600R$.

Two moles of an ideal monoatomic gas undergo a cyclic process as shown in the figure. The temperatures in different states are given as $6 T_1 = 3 T_2 = 2 T_4 = T_3 = 2400 \text{ K}$. The work done by the gas during the complete cycle is $(R = \text{Universal gas constant})$ (in $R$)

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