Question

Calculate the work done in erg for the isothermal reversible expansion of $1mole$ of an ideal gas from volume of $10l$ to $20l$ at ${25}^{o}C$ is $(R=8J{K}^{-1}mo{l}^{-1})$:

• A. $2\times {10}^{10}erg$
• B. $-1.65\times {10}^{10}erg$
• C. $-2.5\times {10}^{4}erg$
• D. $-1.65\times \times {10}^{4}erg$

Answer 1

The correct option is B $-1.65\times {10}^{10}erg$

Work done for isothermal reversible Expansion,
$\Rightarrow W_{rev}=-2.303nRTlog\frac{V_2}{V_1}$
$\Rightarrow W_{rev}=-2.303\times 1\times 8\times 298log\frac{20}{10}\text{J}$
$\Rightarrow W_{rev}=-2.303\times 8\times 298\times log2\text{J}$
$\Rightarrow W_{rev}=-2.303\times 8\times {10}^7\times 298\times log2erg$
$\Rightarrow W_{rev}=-1.65\times {10}^{10}erg$