Question

Two moles of an ideal gas initially at ${27}^{0}C$ and $50L$ volume are compressed isothermally and reversibly till the final volume of the gas is $5L$. Calculate $q$ for the process.

• A. $-11.4kJ$
• B. $-25.4kJ$
• C. $-30.1kJ$
• D. $-21.4kJ$

Answer 1

The correct option is A $-11.4kJ$

work done in an isothermal reversible process is given as:

$w=-2.303nRT\left(\log \frac{V_2}{V_1}\right)$
Here, $n=2mol$
$T=27+273=300K$
$V_1=50L,V_2=5atm$
$R=8.314J{K}^{-1}mo{l}^{-1}$

Using the formula we get,
$w=-2.303\times 2\times 8.314\times 300\times \log \frac{5}{50}$
$w=11488J$

For isothermal process, $△U=0$ (as temperature is constant)
so, from first law
$△U=q+w$
$q=-w=-11488J$
or
$q=-11.4kJ$