Question

A gas in a closed container is heated from $0^{o}Cto{273}^{o}C$. How does this affect the pressure of the gas on the walls of the container?

Answer 1

We know by Gay Lussac's law that volume remaining consant , the pressure of the given mass of a gas is directly proportional to its Kelvin temperature ,

$P\propto T$ ,

if $P_1=$ initial pressure of a gas ,

$P_2=$ final pressure of gas ,

$T_1=$ initial temperature of gas ,

$T_2=$ final temperature of gas , ,

then $P_2/P_1=T_2/T_1$ ,

given $T_1=0+273=273K$ , $T_2=273+273=546K$ ,

and volume is constant because gas is in a closed container ,

therefore $P_2/P_1=546/273=2$ ,

$P_2=2P_1$ ,

pressure of gas will be doubled.