Question

The pressure of a given mass of gas at 27°C is
75 cm of mercury. Find the temperature in °C
at which the pressure is doubled, the gas being
heated at constant volume.​

Answer 1

Answer:

327°C

Explanation:

According to Gay-Lussac's law:

At constant Volume, Pressure (in atm) ∝ Temperature (in K) ,

Therefore, $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Here, $P_1$ = 75 cmHg ≅ 1 atm;

         $T_1=$ 27°C ≅ 300 K

Since  Pressure is doubled, $P_2 =$ 2 atm

We get,

             $\frac{1}{300} = \frac{2}{T_2}$

       ⇒  $T_2 = 600K = 327$ °C