Question

If the pressure of the gas contained in a closed vessel is increased by $40\%$ when heated by $275\text{K}$ then, its initial temperature must have been:

• A. $1375\text{K}$
• B. $345.6\text{K}$
• C. $687.5\text{K}$
• D. $1000.5\text{K}$

Answer 1

The correct option is C $687.5\text{K}$

Let,
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 $T_2=(T+275)\text{K}$
Final Pressure of the gas $P_2=P+0.4P\text{atm}$
From Gay Lussac's law, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ at constant volume.

$\frac{P}{T}=\frac{1.4P}{T+275}$

$\Rightarrow 1.4T=T+275\Rightarrow 0.4T=275$

$\Rightarrow T=687.5\text{K}$