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An open bulb containing air at $29^{0} C$ was cooled to a certain temperature at which the number of moles of the gaseous molecules increased by 50%. The final temperature is

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$233.6 K$

$240 K$

$201.3 K$

$39.6 K$

answer is C.

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Detailed Solution

$n_{1} T_{1} = n_{2} T_{2}$

$n_{1} = (\frac{x}{m}) T_{1} = 302 K$

$n_{2} = \frac{x}{M} + \frac{x}{2 M} = \frac{3 x}{2 M}$ $T_{2} = ?$

$\frac{x}{M} \times 0.302 = \frac{3 x}{2 M} \times T_{2} \Rightarrow \frac{600}{3} = 201.3 K$

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