Question

A container of initial volume $0.15 \text{ m}^3$ is expanded adiabatically from pressure 8 bar to final pressure 1 bar. If initial temperature is $140 \text{ K}$, then find work done by the gas :- ($C_P = 3R, C_V = 2R$) :-

Answer 1

Step 1: Find the adiabatic exponent $\gamma$.
$\gamma = \frac{C_P}{C_V} = \frac{3R}{2R} = 1.5$.

Step 2: Find final volume $V_2$ using adiabatic relation $P_1 V_1^\gamma = P_2 V_2^\gamma$.
$V_2 = V_1 \left( \frac{P_1}{P_2} \right)^{1/\gamma} = 0.15 \times \left( \frac{8}{1} \right)^{1/1.5} = 0.15 \times (8)^{2/3}$.
$(8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$.
$V_2 = 0.15 \times 4 = 0.6 \text{ m}^3$.

Step 3: Calculate work done in adiabatic process: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
Note: $1 \text{ bar} = 10^5 \text{ Pa}$.
$P_1 V_1 = (8 \times 10^5) \times 0.15 = 1.2 \times 10^5 \text{ J}$.
$P_2 V_2 = (1 \times 10^5) \times 0.6 = 0.6 \times 10^5 \text{ J}$.
$W = \frac{(1.2 - 0.6) \times 10^5}{1.5 - 1} = \frac{0.6 \times 10^5}{0.5} = 1.2 \times 10^5 \text{ J} = 120 \text{ KJ}$.

Final Answer: 120 KJ.