Question

One mole of ideal gas undergoes cyclic process shown in figure. Process 1®2 is adiabatic. Heat given to gas in the process is (nearly) –

• A. – 59 J
• B. – 47 J
• C. – 42 J
• D. None of these

Answer 1

Answer: (B)

– 47 J

Answer 2

W₁₂ = –$\frac{nR(T_{1} - T_{2})}{\gamma - 1}$= –$\frac{RT_{1}(10^{\frac{\gamma - 1}{\gamma}} - 1)}{\gamma - 1}$

= – 5600 kJ

W₂₃ = 0

W₃₁ = nRT₁ ln $\left( \frac{V}{V/10} \right)$ = RT₁ ln 10 = 5552.67 J

\ DQ = W_net = – 47.33 J