Question

A gas cylinder containing cooking gas can withstand a pressure 14.9 atmosphere. The pressure guage of the cylinder indicates 12 atmosphere at $27{}^{\circ }\text{C}$. Due to a sudden fire in the building, the temperature starts rising. Find the temperature ( K) when the cylinder explodes?

Answer 1

The cylinder can withstand a maximum pressure of 14.9 atm. So at this pressure, the cylinder explodes and this can be taken as the final pressure.
Given information:
Gauge pressure = 12 atm
Atmospheric pressure = 1 atm
Initial pressure of the gas in cylinder $P_1=12+1=13$ atm
Initial temperature of the gas $T_1=(27+273)=300K$
Final pressure of the gas $P_2=14.9atm$
Final temperature of the gas $T_2=?$
From Gay lussac's law, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ at constant volume.

$\Rightarrow T_2=P_2{T}_1/P_1=\frac{14.9\times 300}{13}=343.8K$

Theory:

Gay lussac’s law :

For a fixed amount of gas at constant volume, pressure of the gas is directly proportional to temperature of the gas on the absolute scale of temperature.

$P\propto Temperature\left(Absolute\right)$

Absolute temperature is temperature in kelvin(K).

Experiment :

Cylinder filled with some gas heated constantly is covered with a piston from top.Pressure in the gas increases by the increase in temperature which is balanced by the weight added on the piston constantly by sprinkling sand such that the volume of the cylinder remains constant.In the process internal pressure of the gas increases at constant volume due to increase in temperature of the gas.

According to the Gay Lusssac’s law :

For a given amount of gas at constant volume :

$P\propto T$

$P=k_{3}T$,$\frac{P}{T}=k_3\left(constant\right)$

Here $k_3$ depends on the amount of gas and volume of the gas.

For a given amount of gas at constant volume changes to a different pressure and temperature :

$\frac{P_i}{T_i}=\frac{P_f}{T_f}$

Where $P_{i}and{T}_i$ are initial pressure and temperature and $P_{f}and{T}_f$ are final pressure and temperature.