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(A) The work done by the gas in a cyclic process on a $P-V$ diagram is equal to the area enclosed by the cycle.For the given process $1-2-3-4-1$, the

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$1650$

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(A) The work done by the gas in a cyclic process on a $P-V$ diagram is equal to the area enclosed by the cycle.For the given process $1-2-3-4-1$, the area is the area of the trapezoid formed by the points $(P_2, V_2), (P_2, V_3), (P_1, V_4), (P_1, V_1)$.Since the lines $1-2$ and $3-4$ pass through the origin, they represent processes where $P \propto V$, meaning $P/V = 	ext{constant}$.Using the ideal gas equation $PV = nRT$, we have $P = (nR/V)T$. Since $P/V$ is constant, $T/V^2$ is constant.However, it is simpler to calculate the work done as the sum of work in individual processes:$W_{1-2} = 0$ (since it is a line through the origin, $P/V = k$, so $W = \int P dV = \int kV dV = \frac{k}{2}(V_2^2 - V_1^2) = \frac{1}{2}(P_2 V_2 - P_1 V_1) = \frac{nR}{2}(T_2 - T_1)$).$W_{2-3} = P_2(V_3 - V_2) = nR(T_3 - T_2)$.$W_{3-4} = 0$ (similar to $1-2$, $W = \frac{nR}{2}(T_4 - T_3)$).$W_{4-1} = P_1(V_1 - V_4) = nR(T_1 - T_4)$.Total Work $W = W_{1-2} + W_{2-3} + W_{3-4} + W_{4-1} = \frac{nR}{2}(T_2 - T_1) + nR(T_3 - T_2) + \frac{nR}{2}(T_4 - T_3) + nR(T_1 - T_4)$.$W = nR [ \frac{1}{2}T_2 - \frac{1}{2}T_1 + T_3 - T_2 + \frac{1}{2}T_4 - \frac{1}{2}T_3 + T_1 - T_4 ]$.$W = nR [ \frac{1}{2}T_1 - \frac{1}{2}T_2 + \frac{1}{2}T_3 - \frac{1}{2}T_4 ] = \frac{nR}{2} (T_1 - T_2 + T_3 - T_4)$.Given $n = 3, T_1 = 400 \ K, T_2 = 700 \ K, T_3 = 2500 \ K, T_4 = 1100 \ K$.$W = \frac{3R}{2} (400 - 700 + 2500 - 1100) = \frac{3R}{2} (1100) = 1650 R$.

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