Question

One mole of an ideal gas expands reversibly at a constant temperature of $300\mathrm{K}$ from an initial volume of $10\mathrm{L}$ to a final volume of $20\mathrm{L}$. What will be the work done in expanding the gas?

• A. $750\mathrm{J}$
• B. $1500\mathrm{J}$
• C. $1728\mathrm{J}$
• D. $3456\mathrm{J}$

Answer 1

The correct option is C

$1728\mathrm{J}$

Explanation for Correct Answer:

Option C $1728\mathrm{J}$

Step 1 Given data:

Temperature, $\mathrm{T}=300\mathrm{K}$

Volume, ${\mathrm{V}}_1=10\mathrm{L}$

Volume, ${\mathrm{V}}_2=20\mathrm{L}$

Step 2 Work done:

Here, one mole of an ideal gas expands reversibly at a constant temperature.

For, the isothermal process, the work done is given by,

$$\begin{gathered} \mathrm{W}=\mu {\mathrm{RTlog}}_{\mathrm{e}}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1} \\ \mathrm{W}=1\times 8.31\times 300{\log }_{\mathrm{e}}\frac{20}{10} \\ \mathrm{W}=1728\mathrm{J} \end{gathered}$$

Therefore, the work done for one mole of an ideal gas expands reversibly at a constant temperature is $1728\mathrm{J}$.