Question

A gas has its pressure decrease from 760 torr to 320 torr at 225k. Was is the initial temperature of the gas?

Answer 1

`T_(1)= 94.75 K`

Step-by-step explanation:

Given the following data;

Final pressure, P2 = 760 Torr

Initial pressure, P1 = 320 Torr

Final temperature, T2 = 225K

To find the initial temperature, we would use Gay Lussac's law;

Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

`PT = K`

`(P1)/(T1) = (P_(2))/(T_(2))`

Making T1 as the subject formula, we have;

`T_(1)= (P_(1))/(P_(2)) * T_(2)`

Substituting into the equation, we have;

`T_(1)= (320)/(760) * 225`

`T_(1)= 0.4211 * 225`

`T_(1)= 94.75 K`

Therefore, the initial temperature of the gas is 94.75 Kelvin.