Question

An LPG cylinder can withstand pressure difference of 15 atm across its boundaries. If at room temp $\left({27}^o\right)$ it is filled with 2 atm pressure. Determine the temperature at which it will explode?

• A. ${216}^{o}C$
• B. ${2127}^{o}C$
• C. ${2400}^{o}C$
• D. ${27}^{o}C$

Answer 1

The correct option is B ${2127}^{o}C$

Given,

Pressure difference across LPG cylinder boundaries= 15 atm

We know atmospheric pressure = 1 atm

It is given that pressure difference across wall =15 atm

therefore,Pressure inside cylinder's wall=15+1=16 atm

$P_1=16atm$

$P_2=2atm$

$T_2=300K$

Now according to gay lusaac's law at constant $V$

$P\propto T$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

$T_1=16\times 300/2$

$T_1=2400K$