Question

Calculate $q$ (heat transfered) for an isothermal reversible expansion of $1mole$ of an ideal gas from an initial pressure of $1.0bar$ to a final pressure of $0.1bar$ at a constant temperature of $273K$.

• A. $7kJ$
• B. $5.2kJ$
• C. $4.3kJ$
• D. $6kJ$

Answer 1

The correct option is B $5.2kJ$

As in isothermal process as temperature remains constant so $△U$ is zero (as it is a function of temperature only)
$△U$ = 0

According to first law of thermodynamics,
$△U=W+q$
$0=W+q$
$q=-W$
as $W=-2.303nRT\left(\log \frac{P_1}{P_2}\right)$
$q=-W=2.303nRT\left(\log \frac{P_1}{P_2}\right)$
Given : $P_1=1bar$
$P_2=0.1bar$
$T=273K$
$n=1mol$

$q=2.303\times 1\times 8.314\times 273\left(\log \frac{1}{0.1}\right)$
$=5227.1Jq=5.2kJ$