A glass bulb contains air at one-atmosphere pressure and $27^{o} C$ . The maximum pressure which the bulb can withstand is $2$ atmosphere. Find the temperature at which it bursts (in $K$ ).

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Solution

According to Gay Lussac's law, the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature, i.e.,
$P \propto T$
$∴ \frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}$
Let $T$ be the temperature at which the bulb burst.
Here,
$P_{1} = 1 a t m$
$P_{2} = 2 a t m$
$T_{1} = 27 ℃ = (27 + 273) K = 300 K$
$T_{2} = ?$
$∴ \frac{1}{2} = 300 T$
$\Rightarrow T = 300 \times 2 = 600 K$
Hence the temperature at which the bulb bursts is $600 K$ .

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A glass bulb contains air at one-atmosphere pressure and 27oC. The maximum pressure which the bulb can withstand is 2 atmosphere. Find the temperature at which it bursts (in K).

A glass bulb contains air at one-atmosphere pressure and $27^{o} C$ . The maximum pressure which the bulb can withstand is $2$ atmosphere. Find the temperature at which it bursts (in $K$ ).