Question

Calculate the work done when 1 mol of an ideal gas undergoes expansion reversibly from $20.0d{m}^3$ to $40.0d{m}^3$ at a constant temperature of 300 K.

• A. $-14.01kJ$
• B. $+18.02kJ$
• C. $-1.72kJ$
• D. $-8.02kJ$

Answer 1

The correct option is C $-1.72kJ$

Solution:-

As the temperature remains constant, thus the process is isothermal.

Work done in an isothermal reversible expansion is given as-

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

Given:-

$n=1\text{mol}$

$V_i=20{dm}^3$

$V_f=40{dm}^3$

$T=300K$

$\therefore W=-2.303\times 1\times 8.314\times 300\times \log \frac{40}{20}$

$\Rightarrow W=-1728.98J=-1.7kJ$