Question

The workdone by a gas of one mole at constant temperature of $27^\circ\text{C}$ when its volume doubled is (Take $\log_{10} 2 = 0.3010$ and Universal gas constant $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$):

• A. $1059\text{ J}$
• B. $1729\text{ J}$
• C. $1679\text{ J}$
• D. $865\text{ J}$

Hint:

For isothermal expansion, remember that $\ln 2 \approx 0.693$.
The work formula simplifies to $W \approx 0.693 \times R T$.
Here, $R T = 8.314 \times 300 = 2494.2\text{ J}$.
Multiplying $2494.2 \times 0.693 \approx 1728.5\text{ J}$, which immediately points to $1729\text{ J}$.

Answer 1

The Correct Option is B

Step 1: Convert the log value to a natural log up front.
Instead of carrying the $2.303$ conversion factor through the whole calculation, first turn $\log_{10}2=0.3010$ directly into $\ln2$: \[ \ln2 = 2.303\times0.3010 = 0.6934 \]
Step 2: Use the isothermal work formula with this natural log value. \[ W = nRT\ln\left(\frac{V_2}{V_1}\right) = nRT\ln2 \]
Step 3: Plug in the numbers.
With $n=1$, $R=8.314\text{ J mol}^{-1}\text{K}^{-1}$, and $T=300\text{ K}$, \[ W = 1\times8.314\times300\times0.6934 \approx 1729\text{ J} \]
\[ \boxed{W \approx 1729\text{ J}} \]