Question

A cylinder containing cooking gas can withstand a pressure of $18atm$. The pressure gauge of cylinder indicates $12atm$ at ${27}^{o}C$. At what temperature cylinder may explode?

• A. ${77}^{o}C$
• B. $177K$
• C. ${177}^{o}C$
• D. $254K$

Answer 1

Answer: (C)

${177}^{o}C$

Here answer is option $C$.

Since the gas is confirmed in a cylinder, its volume will remain constant.
From the given data,
$P_1=12atm{T}_1=27+273=300K{P}_2=18atm{T}_2=?$
Applying pressure temperature law,
$\frac{P_1}{P_2}=\frac{T_1}{T_2}$
or $T_2=\frac{T_1\times P_2}{P_1}$
Substituting the values in the above equation,
$T_2=\frac{300\times 18}{12}=450K$
Temperature in ${}^{o}C=450-273$
$=177{}^{o}C$
Hence the cylinder will be explode at ${177}^{o}C$