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Q.
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 $25 \,{}^° C$?
Thermodynamics
Solution:
As we know that, work done in isothermal expansion is given as-
$W =- nRT \ln \frac{ V _{2}}{ V _{1}}=-2.303 nRT \log \frac{ V _{2}}{ V _{1}}$
Given:-
$n=$ No. of moles $=1$ mole
$R =$ Gas constant $=2$ cal
$T =$ constant temperature associated with the process
$=25^{\circ} C =(25+273) K =298\, K$
$V _{1}=$ Initial volume $=15\, L$
$V _{2}=$ Final volume $=50\, L$
$\therefore W =-2.303 \times 1 \times 2 \times 298 \times \log \left(\frac{50}{15}\right)$
$\Rightarrow W=-1372.588 \times(\log 10-\log 3)$
$\Rightarrow W =-1372.588 \times 0.523=-717.86 \approx=-718 cal$
Hence the work done will be $-718$ cal.