Question:

An ideal gas expands from volume $V_{1}$ to $V_{2}$ through an isothermal reversible process. The work done by the system is given by

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In standard IUPAC convention, expansion work is negative because energy leaves the system, giving us the signature negative sign: $-nRT \ln(V_{2}/V_{1})$.
Updated On: Jun 3, 2026
  • $W = -nRT \ln\left(\frac{V_{2}}{V_{1}}\right)$
  • $W = nRT \ln\left(\frac{V_{2}}{V_{1}}\right)$
  • $W = zero$
  • $W = -P\Delta V$
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The Correct Option is A

Solution and Explanation

Step 1: Concept
In thermodynamics, the work done during a reversible transformation is calculated by integrating the expression $W = -\int P \, dV$.

Step 2: Meaning
For an ideal gas equation, we substitute $P = \frac{nRT}{V}$ into the integral since the process is isothermal ($T$ remains constant).

Step 3: Analysis
Performing the definite integration from the initial volume $V_{1}$ to the final volume $V_{2}$: $$W = -\int_{V_{1}}^{V_{2}} \frac{nRT}{V} \, dV = -nRT \int_{V_{1}}^{V_{2}} \frac{1}{V} \, dV$$ $$W = -nRT [\ln V]_{V_{1}}^{V_{2}} = -nRT \ln\left(\frac{V_{2}}{V_{1}}\right)$$

Step 4: Conclusion
This matches the standard thermodynamic representation for reversible isothermal work done by an expanding system.

Final Answer: (A)
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