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)

For an ideal gas undergoing a reversible adiabatic (isentropic) process the two standard relations are
  • $$P_1V_1^{\gamma}=P_2V_2^{\gamma}$$
  • $$W=\dfrac{P_2V_2-P_1V_1}{1-\gamma}$$

Step 1 : Convert the given data to SI units
Initial pressure, $$P_1 = 8\text{ bar}=8\times10^{5}\,\text{Pa}$$
Final pressure, $$P_2 = 1\text{ bar}=1\times10^{5}\,\text{Pa}$$
Initial volume, $$V_1 = 0.15\,\text{m}^3$$
Initial temperature, $$T_1 = 140^{\circ}\text{C}=140+273 = 413\,\text{K}$$

Step 2 : Calculate the ratio of specific heats
Given $$c_p = 3R,\;c_v = 2R\;$$ so $$\gamma=\dfrac{c_p}{c_v}=\dfrac{3R}{2R}=1.5=\dfrac{3}{2}$$

Step 3 : Obtain the final volume $$V_2$$ using $$P_1V_1^{\gamma}=P_2V_2^{\gamma}$$
$$V_2 = V_1\left(\dfrac{P_1}{P_2}\right)^{\frac{1}{\gamma}} = 0.15\left(\dfrac{8}{1}\right)^{\frac{1}{1.5}} = 0.15\,(8)^{\frac{2}{3}} = 0.15 \times 4 = 0.60\,\text{m}^3$$

Step 4 : Compute the work done
Using $$W=\dfrac{P_2V_2-P_1V_1}{1-\gamma}$$ with $$\gamma=1.5$$:

$$P_2V_2 = (1\times10^{5})(0.60)= 6.0\times10^{4}\,\text{J}$$
$$P_1V_1 = (8\times10^{5})(0.15)=1.20\times10^{5}\,\text{J}$$

$$W = \dfrac{6.0\times10^{4}-1.20\times10^{5}}{1-1.5} = \dfrac{-6.0\times10^{4}}{-0.5} = 1.20\times10^{5}\,\text{J} = 120\,\text{kJ}$$

Work done during the adiabatic expansion = 120 kJ.