Question

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

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

Answer 1

Answer: (C)

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 $=1$ mole
$R=$ Gas constant $=2$ cal
$T=$ constant temperature associated with the process
$=25^{\circ }C=(25+273)K=298K$
$V_1=$ Initial volume $=15L$
$V_2=$ Final volume $=50L$
$\therefore W=-2.303\times 1\times 2\times 298\times \log \left(\frac{50}{15}\right)$
$\Rightarrow W=-1372.588\times (\log 10-\log 3)$
$\Rightarrow W=-1372.588\times 0.523=-717.86\approx =-718cal$
Hence the work done will be $-718$ cal.