Question:
Calculate the work done when \(1\) mole of an ideal gas is expanded reversibly and isothermally from initial pressure \(10\) bar to final pressure \(1\) bar at constant temperature \(300\) K. [\(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)]
Calculate the work done when \(1\) mole of an ideal gas is expanded reversibly and isothermally from initial pressure \(10\) bar to final pressure \(1\) bar at constant temperature \(300\) K. [\(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)]
Show Hint
Use $w=-2.303\,nRT\log\frac{P_1}{P_2}$ for reversible isothermal expansion.
Updated On: Oct 1, 2026
- \(-5.744\) kJ
- \(-5.123\) kJ
- \(-6.514\) kJ
- \(-4.981\) kJ
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The Correct Option is A
Solution and Explanation
Step 1: Write the formula
For a reversible isothermal expansion, \(w=-2.303\,nRT\log_{10}\frac{V_2}{V_1}\). At constant \(T\), \(\frac{V_2}{V_1}=\frac{P_1}{P_2}\).
Step 2: Substitute
\(n=1\), \(R=8.314\), \(T=300\) K, \(\frac{P_1}{P_2}=10\), so \(\log10=1\).
\[ w=-2.303\times8.314\times300\times1=-5744\text{ J} \]
Step 3: Convert
\(w=-5.744\) kJ. The negative sign means the gas does work on the surroundings. Option (A).
Final Answer:
Work done is \(-5.744\) kJ, option (A). \[ \boxed{\text{(A) } -5.744\text{ kJ}} \]
For a reversible isothermal expansion, \(w=-2.303\,nRT\log_{10}\frac{V_2}{V_1}\). At constant \(T\), \(\frac{V_2}{V_1}=\frac{P_1}{P_2}\).
Step 2: Substitute
\(n=1\), \(R=8.314\), \(T=300\) K, \(\frac{P_1}{P_2}=10\), so \(\log10=1\).
\[ w=-2.303\times8.314\times300\times1=-5744\text{ J} \]
Step 3: Convert
\(w=-5.744\) kJ. The negative sign means the gas does work on the surroundings. Option (A).
Final Answer:
Work done is \(-5.744\) kJ, option (A). \[ \boxed{\text{(A) } -5.744\text{ kJ}} \]
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