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

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

Step 1: Calculate \( \gamma \)
The ratio \( \gamma \) is given by: \[ \gamma = \frac{C_P}{C_V} = \frac{3R}{2R} = 1.5. \]

Step 2: Use the adiabatic relation
The adiabatic relation between pressure and volume is: \[ P_1 V_1^\gamma = P_2 V_2^\gamma. \] From this, we can find \( V_2 \) in terms of \( V_1 \): \[ \frac{V_2}{V_1} = \left( \frac{P_1}{P_2} \right)^{\frac{1}{\gamma}}. \] Given: \[ P_1 = 8 \, \text{bar} = 8 \times 10^5 \, \text{Pa}, \quad P_2 = 1 \, \text{bar} = 10^5 \, \text{Pa}. \] Substituting these values: \[ \frac{V_2}{V_1} = \left( \frac{8 \times 10^5}{10^5} \right)^{\frac{1}{1.5}} = 8^{\frac{1}{1.5}} \approx 4. \] Thus, \( V_2 = 4 V_1 \).

Step 3: Use the first law of thermodynamics
The work done during an adiabatic expansion is given by: \[ W = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma}. \] Substituting \( \gamma = 1.5 \), \( P_2 = 10^5 \, \text{Pa} \), \( P_1 = 8 \times 10^5 \, \text{Pa} \), and \( V_2 = 4V_1 \): \[ W = \frac{(10^5 \, \text{Pa}) (4 V_1) - (8 \times 10^5 \, \text{Pa}) (V_1)}{1 - 1.5}. \] Simplifying the numerator: \[ W = \frac{(4 \times 10^5) V_1 - (8 \times 10^5) V_1}{-0.5} = \frac{-4 \times 10^5 V_1}{-0.5} = 8 \times 10^5 V_1. \]

Step 4: Find \( V_1 \) using the ideal gas law
The ideal gas law states: \[ P_1 V_1 = n R T_1, \] where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T_1 = 140 \, \text{K} \). From this, we can solve for \( V_1 \): \[ V_1 = \frac{n R T_1}{P_1}. \] Given \( P_1 = 8 \times 10^5 \, \text{Pa} \) and \( T_1 = 140 \, \text{K} \), we can express the work done as: \[ W = 8 \times 10^5 \times \frac{n R T_1}{P_1}. \]

Step 5: Final answer
After simplifying the equation and performing the necessary calculations, we find that the work done by the gas is: \[ W = 120 \, \text{kJ}. \]

Final Answer: The work done by the gas is \( 120 \, \text{kJ} \).