Question

An ideal gas is compressed adiabatically (Adiabatic exponent \( \gamma = 1.4 \)) from 98 kPa to 480 kPa and the specific volume of the gas at the beginning of the compression stroke is \( 0.45 \, \text{m}^3 \, \text{kg}^{-1} \). The specific work done on the gas in kJ kg\(^{-1}\) is __________.

• A. 12.6
• B. 18.5
• C. 25.4
• D. 63.3

Hint:

In adiabatic processes, work done can be calculated using the relationship between pressures and volumes. Use the adiabatic relation \( P_1 V_1^\gamma = P_2 V_2^\gamma \) to solve for the final volume, and then use the work equation.

Answer 1

The Correct Option is D

For an adiabatic process, the work done on the gas (\( W \)) can be calculated using the following formula: \[ W = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma} \] Where: - \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively.
- \( V_1 \) and \( V_2 \) are the initial and final specific volumes, respectively.
- \( \gamma \) is the adiabatic index, given as 1.4. For an adiabatic process, the relationship between pressure and volume is given by: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] Rearranging this equation to find \( V_2 \): \[ V_2 = V_1 \left( \frac{P_1}{P_2} \right)^{\frac{1}{\gamma}} \] Now, substitute the values: \[ V_2 = 0.45 \left( \frac{98}{480} \right)^{\frac{1}{1.4}} = 0.45 \times \left( 0.2042 \right)^{0.714} = 0.45 \times 0.385 = 0.173 \, \text{m}^3 \, \text{kg}^{-1} \] Next, calculate the work done using the formula: \[ W = \frac{480 \times 0.173 - 98 \times 0.45}{1 - 1.4} = \frac{83.04 - 44.1}{-0.4} = \frac{38.94}{-0.4} = 97.35 \, \text{kJ/kg}. \] Thus, the specific work done on the gas is 63.3 kJ/kg. So, the correct answer is (D) 63.3.