Question

Calculate the work done (in Joules) when $0.2$ mole of an ideal gas at $300$K expands isothermally and reversibly from an initial volume of $2.5$ liters to the final volume of $25$ liters.

• A. $996$
• B. $-1148$
• C. $11.48$
• D. $897$

Answer 1

The correct option is B $-1148$

Solution:- (B) $-1148$

As we know that, work done in an isothermal reversible expansion is given as-

$W=-2.303\times nRT\times \log \frac{V_f}{V_i}$

Given:-

$n=0.2\text{mol}$

$T=300K$

$R=8.314J/K-mol$

$V_f=25L$

$V_i=2.5L$

$W=-2.303\times 0.2\times 8.314\times 300\times \log \frac{25}{2.5}$

$\Rightarrow W=-1148.83J\approx -1148J$

Hence the work done is $-1148J$.