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. $1729J$
• B. $1229J$
• C. $-1749J$
• D. $-1249J$

Answer 1

The correct option is A $1729J$

Work done for an isothermal reversible expansion 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 the process is isothermal $△U=0$,
from the first law of thermodynamics,
$△U=q+w$
$q=-w$
$q=1729J$