Question

a cylinder of gas is at room temperature (20 °C) the air conditioner breaks down and the temperature rises to 40°C what is the new pressure of gas relative to its initial pressure

Answer 1

Given,

The initial temperature of the room, T₁=20 °C=293.15 K

The final temperature, T₂=40 °C=313.15 K

Let the initial temperature of the room be P₁, and the final temperature be P₂

From Gay-Lussac's law,

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

On substituting the known values in the above equation,

$\frac{P_1}{293.15}=\frac{P_2}{313.15}$

Therefore P₂ is given by,

$P_2=\frac{P_1\times313.15}{293.15}=1.07P_1$

Therefore, the new pressure relative to the initial pressure is 1.07 times the initial pressure, i.e., 1.07P₁