A vessel contains 0.15 m ^3 of a gas at pressure 8 bar and temperature 140° C with c_p = 3R and c_v = 2R . It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is _______ kJ. ( R is gas constant)

Step-by-step solution

Given P₁ = 8 bar = 8 10^5 Pa, V₁ = 0.15 m ^3 , and P₂ = 1 bar = 10^5 Pa. The adiabatic index is given by: = c_p c_v = 3R 2R = 1.5 For an adiabatic process, P₁ V₁^ = P₂ V₂^ . V₂ = V₁ ( P₁ P₂ )^ 1 V₂ = 0.15 ( 8 1 )^ 1 1.5 = 0.15 8^ 2 3 = 0.15 4 = 0.6 m ^3 The work done during the adiabatic expansion is: W = P₁ V₁ - P₂ V₂ - 1 W = 8 10^5 0.15 - 1 10^5 0.6 1.5 - 1 W = 1.2 10^5 - 0.6 10^5 0.5 W = 0.6 10^5 0.5 = 1.2 10^5 J W = 120 kJ Answer: 120