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

Nothing can be predicted

m₁ < m₂

m₁ = m₂

m₁ > m₂

Solution

m₁ < m₂

$∵$ pV = nRT

= $\frac{m}{M} RT$

For m₁

p = $\frac{m_{1}}{M} \frac{RT}{V_{1}}$ ....(i)

For m₂

p = $\frac{m_{2}}{M} \frac{RT}{V_{2}}$ .....(ii)

From Eqs. (i) and (ii), we get

$\frac{m_{1}}{M} \frac{RT}{V_{1}} = \frac{m_{2}}{M} \frac{RT}{V_{2}}$

Thus, V ∝ m

$∵$ V₂ > V₁

$∵$ m₂ > m₁

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