Q. Two different isotherms representing the relationship between pressure $p$ and volume $V$ at a given temperature of the same ideal gas are shown for masses $m_{1}$ and $m_{2}$ of the gas respectively in the figure given, then
A) $m_{1} > m_{2}$
B) $m_{1} = m_{2}$
C) $m_{1} < m_{2}$
D) $m_{1}\frac{>}{<}m_{2}$
Explanation
(c) [image:13_image26.emf]
$PV = \mu RT = \frac{m}{M}RT$
For 1st graph,
$P = \frac{m_{1}}{M}\frac{RT}{V_{1}}\ $ …(i)
For 2nd graph,
$P = \frac{m_{2}}{M}\frac{RT}{V_{2}}$ …(ii)
Equating the two, we get, $\frac{m_{1}}{m_{2}} = \frac{V_{1}}{V_{2}} \Rightarrow m \propto V$
As $V_{2} > V_{1} \Rightarrow m_{1} < m_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.