Question

For a reversible adiabatic process involving ideal gas if initial pressure and volume are 8 bar and $0.15 \text{ m}^3$ respectively and final pressure is 1 bar. Calculate |work done| (in Kilo Joule) [$C_V = 2R, C_P = 3R$]

Answer 1

In a reversible adiabatic process for an ideal gas, the relationship between pressure and volume follows $PV^{\gamma} = \text{constant}$, where $\gamma = C_P/C_V$.
1. Find $\gamma$:
$\gamma = \frac{C_P}{C_V} = \frac{3R}{2R} = 1.5 = \frac{3}{2}$

2. Calculate final volume ($V_2$):
$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$
$8 \times (0.15)^{3/2} = 1 \times (V_2)^{3/2}$
Raising both sides to the power $2/3$:
$(8)^{2/3} \times 0.15 = V_2$
$(2^3)^{2/3} \times 0.15 = V_2$
$4 \times 0.15 = V_2 \Rightarrow V_2 = 0.6 \text{ m}^3$

3. Calculate work done (W):
For an adiabatic process:
$W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}$
$W = \frac{(1 \times 0.6) - (8 \times 0.15)}{1.5 - 1}$
$W = \frac{0.6 - 1.2}{0.5} = \frac{-0.6}{0.5} = -1.2 \text{ bar}\cdot\text{m}^3$

4. Convert to Kilo Joules:
$1 \text{ bar}\cdot\text{m}^3 = 10^5 \text{ Pa}\cdot\text{m}^3 = 100 \text{ kJ}$
$W = -1.2 \times 100 \text{ kJ} = -120 \text{ kJ}$
Absolute value $|W| = 120$.