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Ideal Gas Equation

Two different...

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 $m_{2}$ , then

Nothing can be predicted

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m_{1} < m_{2}

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m_{1} = m_{2}

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m_{1} > m_{2}

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Solution

The correct option is B $m_{1} < m_{2}$
$∵ p V = n R T = \frac{m}{M} R T$
For $m_{1}$ , $p = \frac{m_{1}}{M} \cdot \frac{R T}{V_{1}}$ ...(i)
For $m_{2}$ , $p = \frac{m_{2}}{M} \cdot \frac{R T}{V_{2}}$ ....(ii)
From equations (i) and (ii), we get
$\frac{m_{1}}{M} \cdot \frac{R T}{V_{1}} = \frac{m_{2}}{M} \cdot \frac{R T}{V_{2}} \Rightarrow \frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$
Thus, $V \propto m$
$∵ V_{2} > V_{1}$
$∴ m_{2} > m_{1}$

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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 m1 and m2, then

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