Question:
Determine the temperature at which \(2\) moles of oxygen gas expands isothermally and reversibly from \(2\ \text{dm}^{3}\) to \(20\ \text{dm}^{3}\) performs \(-11490\ \text{J}\) of work \([2.303\times R=19.15\ \text{JK}^{-1}\text{mol}^{-1}]\)
Determine the temperature at which \(2\) moles of oxygen gas expands isothermally and reversibly from \(2\ \text{dm}^{3}\) to \(20\ \text{dm}^{3}\) performs \(-11490\ \text{J}\) of work \([2.303\times R=19.15\ \text{JK}^{-1}\text{mol}^{-1}]\)
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Use the reversible isothermal work formula and solve for T.
Updated On: Nov 12, 2026
- \(295\ \text{K}\)
- \(310\ \text{K}\)
- \(305\ \text{K}\)
- \(300\ \text{K}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For a reversible isothermal expansion, \(w = -2.303\,nRT\log\dfrac{V_2}{V_1}\).
Step 2: Substitute
\(\log\dfrac{20}{2} = \log 10 = 1\). So \(-11490 = -2\times 19.15\times T\times 1\).
Step 3: Solve for T
\(T = \dfrac{11490}{38.3} = 300\ \text{K}\).
Step 4: Compare the options
The temperature is \(300\ \text{K}\), the last option.
Final Answer:
The temperature is 300 K. \[ \boxed{\text{(D) }300\ \text{K}} \]
For a reversible isothermal expansion, \(w = -2.303\,nRT\log\dfrac{V_2}{V_1}\).
Step 2: Substitute
\(\log\dfrac{20}{2} = \log 10 = 1\). So \(-11490 = -2\times 19.15\times T\times 1\).
Step 3: Solve for T
\(T = \dfrac{11490}{38.3} = 300\ \text{K}\).
Step 4: Compare the options
The temperature is \(300\ \text{K}\), the last option.
Final Answer:
The temperature is 300 K. \[ \boxed{\text{(D) }300\ \text{K}} \]
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