Question

Calculate the work done when $5$ moles of an ideal gas at $300$K are expanded isothermally from an initial pressure of $4\times {10}^5$ Pa to a final pressure of $1\times {10}^5$ Pa against a constant external pressure of ${10}^5$ Pa. $(5\times 8.314\times 300=12470)$

• A. $-9.35$ kJ
• B. $-16.9$ kJ
• C. Zero
• D. $+16.9$ kJ

Answer 1

The correct option is B $-16.9$ kJ

Solution:- (B) $-16.9kJ$

As the gas is expanded isothermally, thus work done is given by-

$W=-2.303nRT\log \frac{P_i}{P_f}$

Given that:-

$n=5\text{moles}$

$R=8.314J/mol-K$

$T=300K$

$P_i=4\times {10}^{5}Pa$

$P_f={10}^{5}Pa$

$\therefore W=-2.303\times (5\times 8.314\times 300)\log \frac{4\times {10}^5}{1\times {10}^5}$

$\Rightarrow W=-2.303\times 12470\times \log 4(\because 5\times 8.314\times 300=12470)$

$\Rightarrow W=-17.29kJ\approx -16.9kJ$

Hence the work done will be $-16.9kJ$.