Question

Five moles of an ideal gas at ${27}^{o}C$ are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will be:

• A. $-1870.6J$
• B. $-1850.6J$
• C. $-18.47J$
• D. $+18.47J$

Answer 1

The correct option is A $-1870.6J$

Solution:- (A) $-1870.6J$

As we know that,

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

$\Rightarrow W=-P_{ext.}(V_f-V_i)$

From ideal gas equation,

$PV=nRT$

$\Rightarrow V=\frac{nRT}{P}$

Therefore,

$W=-P_{ext.}(\frac{nRT}{P_f}-\frac{nRT}{P_i})$

$\Rightarrow W=-P_{ext.}nRT(\frac{1}{P_i}-\frac{1}{P_f}).....\left(1\right)$

Given:-

External pressure $\left(P_{ext.}\right)=1atm$

Initial pressure $\left(P_i\right)=4atm$

Final pressure $\left(P_f\right)=10atm$

No. of moles $\left(n\right)=5\text{moles}$

Temperature $\left(T\right)=27℃=(27+273)=300K$

Gas constant $\left(R\right)=8.314J/mol-K$

Substituting all these values in ${eq}^n\left(1\right)$, we have

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

$\Rightarrow W=-12471\times \frac{3}{20}=-1870.6J$

Hence the work done will be $-1870.6J$.