Question

Calculate $q$ (in joules), for the reversible isothermal expansion of one mole of an ideal gas ${27}^{o}C$ from a volume of 10 $d{m}^3$ to a volume of $20d{m}^3$.

• A. 1729 J
• B. 1504 J
• C. 1432 J
• D. 1830 J

Answer 1

The correct option is A 1729 J

work is given as
$w=-2.303nRTlog\frac{V_2}{V_1}$

Given : $V_1=10L$
$V_2=20L$
$T={27}^{o}C$ = $27+273=300K$
so,
$w=-1\times 8.314$ $\times 300\times 2.303$ $log\frac{20}{10}$
$w=-1729J$
since process is isothermal $△U=0$
from first law of thermodynamic,
$△U=q+w$
$q=-w$
$q=1729J$