The pressure of a gas in a tank is 3.20 ATM at 22.0°C if the temperature is increased to 60.0°C what will the new pressure in the tank be

P₂ = 3.61 atm

Explanation:

Given data:

Initial pressure of tank = 3.20 atm

Initial temperature = 22.0°C

Final temperature = 60.0°C

Final pressure = ?

Solution:

Initial temperature = 22.0°C (22.0 +273 = 295 K)

Final temperature = 60.0°C (60 +273 = 333)

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:

3.20 atm / 295 K = P₂/333 K

P₂ = 3.20 atm × 333 K / 295 K

P₂ = 1065.6 atm. K /295 K

P₂ = 3.61 atm

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