AP EAPCET 07-Jul-2022 Shift 1 Paper

Process undergone by $1$ mole gas is as shown : For an ideal monoatomic gas, ratio of specific heat $\gamma = \frac{5}{3}$ For adiabatic process $A \rightarrow B$ $p_A V_A^\gamma = p_B V_B^\gamma$ Substituting values of $p_A, p_B$ and $V_B$ from graph, $\Rightarrow 32 \times 10^5 \times V_A^\gamma = 1 \times 10^5 \times (1)^\gamma$ $\Rightarrow V_A^\gamma = \left(\frac{1}{32}\right)^\gamma$ Here, $\gamma = \frac{5}{3}$ $\Rightarrow (V_A)^{\frac{5}{3}} = \frac{1}{32}$ or $V_A = \left(\frac{1}{32}\right)^{\frac{3}{5}} = \frac{1}{(2^5)^{\frac{3}{5}}} = \frac{1}{8} \, \text{m}^3$ Similarly for adiabatic compression $C \rightarrow D$, $p_C V_C^\gamma = p_D V_D^\gamma$ We substitute values from graph, $\Rightarrow (1 \times 10^5) \times (8)^{\frac{5}{3}} = (32 \times 10^5) (V_D)^{\frac{5}{3}}$ $\Rightarrow V_D^{\frac{5}{3}} = 1$ $\Rightarrow V_D = 1^{\frac{3}{5}} = 1 \, \text{m}^3$ Now, Work done in process $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ $= W_{AB} + W_{BC} + W_{CD} + W_{DA}$ Here we substitute work done for adiabatic process, $W_{\text{adiabatic}} = \frac{p_i V_i - p_f V_f}{\gamma - 1}$ and work done of isobaric process, $W_{\text{isobaric}} = p (V_f - V_i)$ so, work done in given cycle $A B C D A$ $= W_{AB} + W_{BC} + W_{CD} + W_{DA}$ $= \frac{p_A V_A - p_B V_B}{\gamma - 1} + p_B (V_C - V_B)$$+ \frac{p_C V_C - p_D V_D}{\gamma - 1} + p_D (V_A - V_D)$ Substituting values of pressures and volumes, we get $W_{ABCDA} = \frac{32 \times 10^5 \times \frac{1}{8} - 1 \times 10^5 \times 1}{\frac{5}{3} - 1}$$+ 1 \times 10^5 (8-1) + \frac{1 \times 10^5 \times 8 - 32 \times 10^5 \times 1}{\frac{5}{3} - 1}$ $+ 32 \times 10^{5} \left(\frac{1}{8} - 1\right)$ $= \frac{3 \times 10^{5} \times 3}{2} + 7 \times 10^{5} + \frac{-24 \times 10^{5} \times 3}{2} + 4 \times 10^{5} \times x - 7 = -52.5 \times 10^{5} \, \text{J}$