Question

If the pressure of a gas contained in a closed vessel is increased by 0.4% when heated by $1{}^{\circ }\text{C}$, its initial temperature must be:

• A. $250\text{K}$
• B. $250{}^{\circ }\text{C}$
  
• C. $25{}^{\circ }\text{C}$
  
• D. $25\text{K}$

Answer 1

The correct option is A $250\text{K}$

Given information:
Initial pressure of the gas $P_1=P\text{atm}$
Initial temperature of the gas $T_1=TK$
Final temperature of the gas $T_2=(T+1)K$
Final Pressure of the gas $P_2=P+0.004P\text{atm}$
From Gay Lussac's law, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ at constant volume.


$\therefore \frac{P}{T}=\frac{1.004P}{T+1}$

$\Rightarrow 1.004T=T+1\Rightarrow 0.004T=1$

$\Rightarrow T=250{}^{0}C$
Hence, Initial temperature $T_1={250}^{0}C$