A gas ocupies a rigid container at a pressure of 54.5 kPa with a temperature of 19.0 °C.
What will the new pressure if the gas is
cooled to -25 °C?

P₂ = 46.3 KPa

Explanation:

Given data:

Initial pressure = 54.5 KPa

Initial temperature = 19.0 °C (19+273 = 292 K)

New pressure = ?

Final temperature = -25°C (-25+273 =248 K)

Solution:

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:

13.0 atm / 293 K = P₂/375 K

P₂ = 54.5 kpa × 248 K / 292 K

P₂ = 13516 KPa. K /292 K

P₂ = 46.3 KPa

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