Q. Work done by 0.1 mole of a gas at $27℃$ to double its volume at constant pressure is $(R = 2\ cal\ mol^{- 1}{^\circ}C^{- 1})$
A) $54\ cal$
B) $600\ cal$
C) $60\ cal$
D) $546\ cal$
Explanation
(c) $\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{V}{2V} = \frac{300}{T_{2}} \Rightarrow T_{2} = 600 \Rightarrow \mathrm{\Delta}T = 300$ Now,
$W = P\mathrm{\Delta}V = \mu R\mathrm{\Delta}T$
$\Rightarrow W = 0.1 \times 2 \times 300 = 60\ cal$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.