Question

The temperature of a certain mass of a gas is doubled. If the initial pressure of the gas is 1 atm, find the % increase in pressure.

• A. 50%
• B. 100%
• C. 150%
• D. 200%

Answer 1

The correct option is B 100%

Given information:
Initial pressure of the gas $P_1=1atm$
Initial temperature of the gas $T_1=TK$
Final temperature of the gas $T_2=2TK$
Final Pressure of the gas $P_2=?$
From Gay Lussac's law, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ at constant volume.

$\frac{1}{T}=\frac{P_2}{2T}$

Hence, $P_2=2atm$

Increase in pressure $=2-1=1$ atm

Percentage increase in pressure $=\frac{\text{increase in pressure}}{\text{initial pressure}}\times 100$$=\frac{1}{1}\times 100=100$ %