Question

Calculate the work done by $0.1$ mole of a gas at $27{}^{o}C$ to double its volume at constant pressure isobaric process $\left(R=2calmo{l}^{-1}{K}^{-1}\right)$

Answer 1

$T={27}^{o}C$
$=27+273=300$
From Ideal gas equation
$PV=nRT$
$w=-pdv$
$=-p\left(2v-v\right)$
$=-p\cdot v\Rightarrow -nRT=0.1mol\times 2calmo{l}^{-1}{k}^{-1}\times 300k$
$=0.1\times 2\times 300=60cal$