The pressure of a gas enclosed in a closed vessel is increased by 0.4% when the temperature is increased by 1𝑜𝐶 . The initial temperature of gas is

The question is based on the principle of Gay-Lussac's law which states that the pressure of a gas is directly proportional to its absolute temperature, provided the volume is kept constant.

The formula for Gay-Lussac's law is P1/T1 = P2/T2.

Given that the pressure of the gas is increased by 0.4%, we can write this as P2 = P1 + 0.004P1 = 1.004P1.

The temperature is increased by 1°C, so T2 = T1 + 1.

Substituting these values into the formula, we get:

P1/T1 = 1.004P1/(T1 + 1)

Solving for T1, we get:

T1 = 1.004P1 * T1 / P1 - 1

Simplifying, we get:

T1 = 1.004T1 - 1

Rearranging, we get:

0.004T1 = 1

Finally, solving for T1, we get:

T1 = 1 / 0.004 = 250°C

So, the initial temperature of the gas is 250°C.