Question

A gas expands from a volume of 2 m³ to 4 m³ against a constant pressure of 5 atm. How much work is done by the gas during expansion? (1 atm = \( 1.01 \times 10^5 \, \text{Pa} \))

• A. \( 2.02 \times 10^5 \, \text{J} \)
• B. \( 1.01 \times 10^5 \, \text{J} \)
• C. \( 5.02 \times 10^5 \, \text{J} \)
• D. \( 1.02 \times 10^5 \, \text{J} \)

Hint:

Remember: The work done during expansion or compression at constant pressure is \( W = P \Delta V \), where \( \Delta V \) is the change in volume.

Answer 1

The Correct Option is A

Given: Initial volume, \( V_1 = 2 \, \text{m}^3 \)
Final volume, \( V_2 = 4 \, \text{m}^3 \)
Constant pressure, \( P = 5 \, \text{atm} = 5.05 \times 10^5 \, \text{Pa} \)

Step 1: Work Done Formula Work done \( W \) at constant pressure is \( W = P \Delta V \), where \( \Delta V = V_2 - V_1 \).

Step 2: Calculate Work Done Using the formula: \( W = (5.05 \times 10^5 \, \text{Pa})(4 \, \text{m}^3 - 2 \, \text{m}^3) = (5.05 \times 10^5)(2) = 1.01 \times 10^6 \, \text{J} \).

Answer: Option (b) is \( 1.01 \times 10^5 \, \text{J} \).