Q. A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 $dm^{3}$ to a volume of $20\ dm^{3}$. It absorbs 800 J of thermal energy from its surroundings. The $\mathrm{\Delta}U$ is
A) $- 312$ J
B) $+ 123$ J
C) $- 213$ J
D) $+ 231$ J
Explanation
(c)
$W = \int_{\frac{1}{1}}^{\frac{1}{2}}{pdV} = - p(V_{2} - V_{1})$
$W = - 1(20 - 10) = - 10\ dm^{3}\ atm$
$= - 10\ dm^{3} \times \frac{8.314\ {JK}^{- 1}{mol}^{- 1}}{0.0821\ {dm}^{3}K^{- 1}{mol}^{- 1}} = - 1013\ J$
From, 1st law of thermodynamics
$\mathrm{\Delta}U = q + W$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 800\ J + ( - 1013\ J) = - 213\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.