Q. The equation of state for 5g of oxygen at a pressure $p$ and temperature $T$, when occupying a volume $V$, will be
A) $pV = \left( \frac{5}{32} \right)RT$
B) $pV = 5RT$
C) $pV = \left( \frac{5}{2} \right)RT$
D) $pV = \left( \frac{5}{16} \right)RT$
Explanation
(a) No. of moles $n = \frac{m}{molecular\ weight} = \ \frac{5}{32}$
So, from ideal gas equation
$pV = nRT$
$\Rightarrow \ \ \ \ \ \ \ \ pV = \frac{5}{32}RT$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.