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

• A. $W = -nRT \ln\left(\frac{V_{2}}{V_{1}}\right)$
• B. $W = nRT \ln\left(\frac{V_{2}}{V_{1}}\right)$
• C. $W = zero$
• D. $W = -P\Delta V$

Hint:

In standard IUPAC convention, expansion work is negative because energy leaves the system, giving us the signature negative sign: $-nRT \ln(V_{2}/V_{1})$.

Answer 1

The Correct Option is A

Step 1: Start from the work formula.
For any reversible change the work done on the gas is \[ W = -\int P \, dV \] The minus sign is the chemistry sign rule, where work done by the gas during expansion comes out negative for the system.

Step 2: Use the gas law.
For an ideal gas, $PV = nRT$, so we can write $P = \dfrac{nRT}{V}$. We put this in place of $P$ inside the integral.

Step 3: Use the isothermal condition.
Isothermal means the temperature $T$ stays fixed. So $n$, $R$ and $T$ are all constants and can be pulled out of the integral.

Step 4: Do the integration.
We integrate from the start volume to the end volume. \[ W = -nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV = -nRT \left[ \ln V \right]_{V_1}^{V_2} \]

Step 5: Put in the limits.
The integral of $1/V$ is the natural log, so \[ W = -nRT \left( \ln V_2 - \ln V_1 \right) = -nRT \ln\left( \frac{V_2}{V_1} \right) \]

Step 6: State the answer.
This is the standard result for reversible isothermal work. \[ \boxed{W = -nRT \ln\left( \frac{V_2}{V_1} \right)} \]