Dear Student, 2T = PVnR=1×202×0 0821=121 80 KTherefore,Now, P1=2 mol×0 0821L-atmmol-1K-1×121 80 K10 L= 2 atmW1= -P1(V2-V1)= -2(10-20)=20 L-atmand, P2=2 mol×0 0821L-atmmol-1K-1×121 80 K2 L= 10 atmW2=-P2(V3-V2)= -10(2-10)=80 L-atmW=W1+W2= 20+80=100 L-attmHence, correct answer is (2)

Q.35

Dear Student,

2

T = \frac{PV}{nR} = \frac{1 \times 20}{2 \times 0.0821} = 121.80 K Therefore, Now, P_{1} = \frac{2 mol \times 0.0821 L - {atmmol}^{- 1} K^{- 1} \times 121.80 K}{10 L} = 2 atm W_{1} = - P_{1} (V_{2} - V_{1}) = - 2 (10 - 20) = 20 L - atm and, P_{2} = \frac{2 mol \times 0.0821 L - {atmmol}^{- 1} K^{- 1} \times 121.80 K}{2 L} = 10 atm W_{2} = - P_{2} (V_{3} - V_{2}) = - 10 (2 - 10) = 80 L - atm W = W_{1} + W_{2} = 20 + 80 = 100 L - attm Hence, correct answer is (2)

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​Q.35

Q.35