Question

$2.8g$ of $N_2$ gas at $300$ K and $20$ atm was allowed to expand isothermally against a constant external pressure of $1$ atm. Calculate $W$ for the gas.

• A. $+236.95$ J
• B. $+136.95$ J
• C. $-236.95$ J
• D. $-136.95$ J

Answer 1

Answer: (C)

$-236.95$ J

Solution:- (C) $-236.95J$

Weight of $N_2=28g$

Therefore,

No. of moles of $N_2\left(n\right)=\frac{2.8}{28}=0.1\text{mole}$

Given:-

$P_{ext.}=1atm$

$T=300K$

As we know that,

$W=-P_{ext.}\Delta V$

$\Rightarrow W=-P_{ext.}(V_2-V_1)$

$\Rightarrow W=-P_{ext.}(\frac{nRT}{P_2}-\frac{nRT}{P_1})[\because V=\frac{nRT}{P}\left(\text{From ideal gas equation}\right)]$

$\Rightarrow W=-P_{ext.}nRT(\frac{1}{P_2}-\frac{1}{P_1})$

$\therefore W=-1\times 0.1\times 8.314\times 300(\frac{1}{1}-\frac{1}{20})$

$\Rightarrow W=-249.42\times \frac{19}{20}=-236.95$

Hence $W$ for the gas is $-236.95J$.