Question

A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of the cylinder indicates 12 atm at ${27^o}C$ . Due to sudden fire in the building, its temperature starts rising. At what temperature cylinder will explode?

Answer 1

In the above situation of a gas cylinder, the volume remains constant, the number of moles of gases in the cylinder will not change, so that is also constant.
Thus, we can apply Gay-Lussac Law of gaseous state, which says- for the fixed amount of gas, at constant volume, pressure of gas is directly proportional to Temperature of that gas.
$P\alpha T$

Complete step by step answer:
The above question has a gas cylinder, and we know that the gas cylinder has a metal container, which has fixed volume, and the amount of gas inside the cylinder will be fixed, that means moles of gases will be fixed.
Now, we know ideal gas equation is:
$PV = nRT$
This is derived from different gas laws, one among these gas laws is Gay-Lussac law.
It states that for the fixed amount of gas, at constant volume, pressure of gas is directly proportional to Temperature of that gas.
$P\alpha T$
Now, we know for the same container and situation, if you have 2 conditions for pressure, then we will get 2 different temperatures corresponding to each pressure. So we can write the relation of pressure and temperature as:
$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$
In question, given data is:

$${P_1} = 12atm \\\ {P_2} = 14.9atm \\\$$

For writing temperature, we need to convert them into Kelvin scale. And we know
C=273+K
Here, C= temperature in Celsius scale, K= temperature in Kelvin scale.

$${T_1} = {27^0}C \\\ = 273 + 27 \\\ = 300K \\\$$

${T_2} = ?$ (we need to find out)
Now when we substitute pressure and temperature values we get,
$\dfrac{{12}}{{14.9}} = \dfrac{{300}}{{{T_2}}}$
Cross multiply and take all numerical values on same side, we get
${T_2} = \dfrac{{300 \times 14.9}}{{12}}$
On simplification, we can get value of temperature in Kelvin scale,
$\therefore {T_2} = 372.5K$
To convert temperature from Kelvin scale into Celsius scale.

$${T_2} = 372.5 - 273 \\\ = {99.5^o}C \\\$$

Thus, the temperature at which cylinder will explode is 372.5 K or ${99.5^o}C$.

Note:
Take care of conversion of temperatures, and even while substituting values into equations, the temperature has to be in Kelvin scale. Take complete note of all gas laws, and when to use which gas law is important, or else even the general ideal gas equation could be used to come to the conclusion of getting an answer of temperature at which the cylinder will explode.