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)

Given \(P_1 = 8\) bar \(= 8 \times 10^5\) Pa, \(V_1 = 0.15\) m\(^3\), and \(P_2 = 1\) bar \(= 10^5\) Pa.

The adiabatic index \(\gamma\) is given by:

\(\gamma = \dfrac{c_p}{c_v} = \dfrac{3R}{2R} = 1.5\)

For an adiabatic process, \(P_1 V_1^\gamma = P_2 V_2^\gamma\).

\(V_2 = V_1 \left( \dfrac{P_1}{P_2} \right)^{\dfrac{1}{\gamma}}\)

\(V_2 = 0.15 \times \left( \dfrac{8}{1} \right)^{\dfrac{1}{1.5}} = 0.15 \times 8^{\dfrac{2}{3}} = 0.15 \times 4 = 0.6\) m\(^3\)

The work done during the adiabatic expansion is:

\(W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}\)

\(W = \dfrac{8 \times 10^5 \times 0.15 - 1 \times 10^5 \times 0.6}{1.5 - 1}\)

\(W = \dfrac{1.2 \times 10^5 - 0.6 \times 10^5}{0.5}\)

\(W = \dfrac{0.6 \times 10^5}{0.5} = 1.2 \times 10^5\) J

\(W = 120\) kJ

Answer: \(120\)