Question

A vessel contains \(0.15\,\text{m}^3\) of a gas at pressure \(8\) bar and temperature \(140^\circ\text{C}\) with \(c_p=3R\) and \(c_v=2R\). It expands adiabatically till pressure falls to \(1\) bar. The work done during this process is ____ kJ. (R is gas constant)}

Answer 1

Correct Answer: 120

Step 1: Understanding the Concept:
Adiabatic expansion follows the relation \(PV^{\gamma} = \text{constant}\). The work done in an adiabatic process is the change in internal energy, given by the formula involving initial and final pressures and volumes.
Step 2: Key Formula or Approach:
1. Adiabatic index: \(\gamma = \frac{c_p}{c_v}\).
2. Adiabatic relation: \(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\).
3. Work done: \(W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}\).
Step 3: Detailed Explanation:
1. Calculate \(\gamma\):
\[ \gamma = \frac{3R}{2R} = 1.5 \]
2. Find final volume \(V_2\):
Given \(P_1 = 8 \text{ bar}, P_2 = 1 \text{ bar}, V_1 = 0.15 \text{ m}^3\).
\[ P_1 V_1^{1.5} = P_2 V_2^{1.5} \implies \left( \frac{V_2}{V_1} \right)^{1.5} = \frac{8}{1} = 8 \]
\[ \frac{V_2}{V_1} = 8^{1/1.5} = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4 \]
\[ V_2 = 4 \times V_1 = 4 \times 0.15 = 0.6 \text{ m}^3 \]
3. Calculate Work Done (\(W\)):
Convert pressure to SI units: \(1 \text{ bar} = 10^5 \text{ Pa}\).
\[ W = \frac{(8 \times 10^5 \times 0.15) - (1 \times 10^5 \times 0.6)}{1.5 - 1} \]
\[ W = \frac{1.2 \times 10^5 - 0.6 \times 10^5}{0.5} = \frac{0.6 \times 10^5}{0.5} = 1.2 \times 10^5 \text{ J} \]
Convert to kJ: \(W = 120 \text{ kJ}\).
Step 4: Final Answer:
The work done is 120 kJ.