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

Thermodynamics of adiabatic expansion. We need to find the final state parameters first, then calculate the work using the adiabatic work formula.

LOGIC:
1. $\gamma = C_P/C_V = 3R/2R = 1.5$.
2. Use $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$ to find $V_2$:
$V_2 = V_1 \left(\frac{P_1}{P_2}\right)^{1/\gamma} = 0.15 \times \left(\frac{8}{1}\right)^{1/1.5} = 0.15 \times (8)^{2/3} = 0.15 \times 4 = 0.6 \text{ m}^3$
3. Calculate work done using temperature change or PV change:
$W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}$
$W = \frac{1 \times 0.6 - 8 \times 0.15}{1.5 - 1} = \frac{0.6 - 1.2}{0.5} = -1.2 \text{ bar}\cdot\text{m}^3$
4. Conversion: $1 \text{ bar}\cdot\text{m}^3 = 10^5 \text{ J} = 100 \text{ kJ}$.
5. Magnitude $|W| = 1.2 \times 100 = 120 \text{ kJ}$.