Question

What will be the work done when one mole of a gas expands isothermally from 15 L to 50 L against a constant pressure of 1 atm at ${25}^0$C?

• A. $-3542cal$
• B. $-843.3cal$
• C. $-718cal$
• D. $-60.23cal$

Answer 1

The correct option is C $-718cal$

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

$W=-nRT\ln \frac{V_2}{V_1}=-2.303nRT\log \frac{V_2}{V_1}$

Given:-

$n=$ No. of moles $=1\text{mole}$

$R=$ Gas constant $=2\text{cal}$

$T=$ constant temperature associated with the process $=25℃=(25+273)K=298K$

$V_1=$ Initial volume $=15L$

$V_2=$ Final volume $=50L$

$\therefore W=-2.303\times 1\times 2\times 298\times \log \left(\frac{50}{15}\right)$

$\Rightarrow W=-1372.588\times (\log 10-\log 3)$

$\Rightarrow W=-1372.588\times 0.523=-717.86\approx =-718\text{cal}$

Hence the work done will be $-718\text{cal}$.