Question

A sample of gas with $\gamma = 1.5$ is taken through an adiabatic process in which the volume is compressed from 1200 $cm^3$ to 300 $cm^3$. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = ________ J.

Hint:

Work done in compression is always negative (work is done on the gas). "Absolute value" indicates the question only requires the magnitude, which is 480.

Answer 1

Correct Answer: 480

Step 1: Understanding the Concept:
In an adiabatic process, the work done is given by the change in internal energy (with opposite sign) or can be calculated directly using the initial and final pressures and volumes.
Step 2: Key Formula or Approach:
1. Adiabatic equation: \( P_1 V_1^\gamma = P_2 V_2^\gamma \).
2. Work done: \( W = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma} \).
Step 3: Detailed Explanation:
Given: \( P_1 = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa} \), \( V_1 = 1200 \times 10^{-6} \text{ m}^3 \), \( V_2 = 300 \times 10^{-6} \text{ m}^3 \), \( \gamma = 1.5 \).
Find \( P_2 \):
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma = 200 \left( \frac{1200}{300} \right)^{1.5} = 200 \times (4)^{3/2} = 200 \times 8 = 1600 \text{ kPa} \]
Calculate Work Done:
\[ W = \frac{(1600 \times 10^3 \times 300 \times 10^{-6}) - (200 \times 10^3 \times 1200 \times 10^{-6})}{1 - 1.5} \]
\[ W = \frac{480 - 240}{-0.5} = \frac{240}{-0.5} = -480 \text{ J} \]
The absolute value is \( |W| = 480 \) J.
Step 4: Final Answer:
The absolute value of the work done is 480 J.