Question

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

• A. $1052{}^{\circ }\text{C}$
• B. $1034{}^{\circ }\text{C}$
• C. ${1092}^{\circ }\text{C}$
• D. $1200\text{K}$

Answer 1

The correct option is C ${1092}^{\circ }\text{C}$

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

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

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

$T=1365K$