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Ideal Gas Equation

An open bulb ...

Question

An open bulb containing air at $19^{0} C$ was cooled to a certain temperature at which the no. of moles of gaseous molecules increased by $25$ %. What is the final temperature?

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Solution

By the ideal gas low:
$P V = n R T$
and acc. to question $n_{2} = 1.25 n_{1}$
$P V / R = n_{1} T_{1}$ and $P V / R = n_{2} T_{2}$
$n_{1} T_{1} = n_{2} T_{2}$
$n_{1} T_{1} = 1.25 n_{1} T_{2}$
$T_{2} = \frac{T_{1}}{1.25} = \frac{292}{1.25} = 233.6 k = 40^{o} C$

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