Question

If the pressure of the gas contained in a closed vessel is increased by $20\%$ when heated by $273K$ then, its initial temperature must have been:

• A. $1052{}^{\circ }\text{C}$
• B. $1029\text{K}$
• C. $1365{}^{\circ }\text{C}$
• D. $1365\text{K}$

Answer 1

The correct option is D $1365\text{K}$

Given information:
Initial pressure of the gas $P_1=P\text{atm}$
Initial temperature of the gas $T_1=T\text{K}$
Final temperature of the gas $T_2=(T+273)\text{K}$
Final Pressure of the gas $P_2=P+0.2P\text{atm}$
From Gay Lussac's law, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ at constant volume.

$\frac{P}{T}=\frac{1.2P}{T+273}$

$1.2T=T+273\Rightarrow 0.2T=273$

$T=1365\text{K}$
Illustration for Gay Lussac's Law: