Question

An ideal gas expands from 2 L to 6 L at a constant temperature of 300 K. Calculate the work done by the gas if pressure remains constant at 2 atm. (1 atm = $1.013 \times 10^5$ Pa)

• A. 810.4
• B. 820
• C. 820.5
• D. 795.5

Hint:

For constant pressure expansion, use $W = P \Delta V$; always convert atm to Pa and L to m\textsuperscript{3} before calculation.

Answer 1

The Correct Option is A

Initial pressure \( P = 2 \, \text{atm} \). This is equivalent to \( 2 \times 1.013 \times 10^5 \, \text{Pa} = 2.026 \times 10^5 \, \text{Pa} \). The change in volume is \( \Delta V = V_f - V_i = 6\, \text{L} - 2\, \text{L} = 4\, \text{L} \). This volume change is \( 4 \times 10^{-3} \, \text{m}^3 \). The work done by the gas at constant pressure is calculated as \( W = P \Delta V \). Using the converted units, \( W = 2.026 \times 10^5 \times 4 \times 10^{-3} = 810.4 \, \text{J} \). However, if the options do not match, let's re-examine the calculation. The work done is \( W = 2 \, \text{atm} \times (6 - 2) \, \text{L} = 2 \times 4 = 8 \, \text{L atm} \). Converting to Joules, \( 1\, \text{L atm} = 101.3 \, \text{J} \), therefore \( W = 8 \times 101.3 = 810.4 \, \text{J} \).