Question:
At 27^C, 1.6 g of O$_2$ gas at 5 atm expands isothermally against a constant external pressure of 1 atm. The work done (in J) is (1 L-atm = 100 J):
At 27^C, 1.6 g of O$_2$ gas at 5 atm expands isothermally against a constant external pressure of 1 atm. The work done (in J) is (1 L-atm = 100 J):
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Always use external pressure for irreversible work calculation: \(W = -P_{\text{ext}}\Delta V\).
Updated On: Jun 10, 2026
- -49.2
- -98.4
- +98.4
- +49.2
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The Correct Option is B
Solution and Explanation
Concept:
For irreversible isothermal expansion against constant external pressure:
\[
W = -P_{\text{ext}}(V_2 - V_1)
\]
Using ideal gas law:
\[
V = \frac{nRT}{P}
\]
Step 1: Moles of gas \[ n = \frac{1.6}{32} = 0.05\,\text{mol} \]
Step 2: Initial volume \[ V_1 = \frac{nRT}{P_1} = \frac{0.05 \times 0.082 \times 300}{5} = 0.246\,\text{L} \]
Step 3: Final volume \[ V_2 = \frac{nRT}{P_{\text{ext}}} = 1.23\,\text{L} \]
Step 4: Work done \[ W = -1 \times (1.23 - 0.246) = -0.984\,\text{L atm} \] \[ W = -0.984 \times 100 = -98.4\,\text{J} \]
Step 1: Moles of gas \[ n = \frac{1.6}{32} = 0.05\,\text{mol} \]
Step 2: Initial volume \[ V_1 = \frac{nRT}{P_1} = \frac{0.05 \times 0.082 \times 300}{5} = 0.246\,\text{L} \]
Step 3: Final volume \[ V_2 = \frac{nRT}{P_{\text{ext}}} = 1.23\,\text{L} \]
Step 4: Work done \[ W = -1 \times (1.23 - 0.246) = -0.984\,\text{L atm} \] \[ W = -0.984 \times 100 = -98.4\,\text{J} \]
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