Question

At constant volume, the pressure of a gas is 755 kPa at 30.0 degrees C. What is the temperature of the gas if the pressure is decreased to 252 kPa?

Answer 1

T₂ = 101.13 K OR

- 172.02 °C

Step-by-step explanation:

Given data:

Initial pressure of gas = 755 Kpa

Initial temperature = 30.0°C

Final temperature = ?

Final pressure = 252 Kpa

Solution;

Initial temperature = 30.0°C (30+273 = 303 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

755 Kpa / 303 K = 252 Kpa / T₂

T₂ = 252 Kpa ×303 K / 755 Kpa

T₂ = 76356 KPa.K / 755 Kpa

T₂ = 101.13 K

Kelvin to °C:

101.13 K - 273.15 = - 172.02 °C