Question

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. $m_{1} \geq m_{2}$

Answer 1

Answer: (C)

$p V=n R T=\frac{m}{M} R T$

For $1 ^\text{st}$ graph, $p=\frac{m_{1}}{M} \frac{R T}{V_{1}}$
For $2^\text{ nd}$ graph, $p=\frac{m_{2}}{M} \frac{R T}{V_{2}}$
Equating the two, we get
$\frac{m_{1}}{m_{2}}=\frac{V_{1}}{V_{2}} $
$\Rightarrow m \propto \frac{1}{V}$
As $V_2 > V_1$
$\Rightarrow m_1 < m_2$