Question

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0∘C. What is the new absolute pressure p2? Express your answer in terms of p1.

Answer 1

Answer : The new absolute pressure is, $1.068\times P_1$

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

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

where,

$P_1$ = initial pressure of gas

$P_2$ = final pressure of gas

$T_1$ = initial temperature of gas = $20.0^oC=273+20.0=293.0K$

$T_2$ = final temperature of gas = $40.0^oC=273+40.0=313.0K$

Now put all the given values in the above equation, we get:

$\frac{P_2}{P_1}=\frac{313.0K}{293.0K}$

$\frac{P_2}{P_1}=1.068$

$P_2=1.068\times P_1$

Therefore, the new absolute pressure is, $1.068\times P_1$