Problem: A flask contains air at 722 mm Hg and 22°C. What would the temperature of the gas be if the pressure is increased to 1.07 atm?

FREE Expert Solution

Gay Lussac’s Law:


P1T1=P2T2


We can use Gay Lussac’s law as long as the units for the initial and final values are the same.


The given values are:

P1 = 722 mmHg

P1=722 mmHg×1 atm760 mmHg

P2 = 0.95 atm


P2 = 1.07 atm

T1 = 22°C +273.15 = 295.15 K              

T2 = ??


We can rearrange the equation to solve for T2:

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Problem Details

A flask contains air at 722 mm Hg and 22°C. What would the temperature of the gas be if the pressure is increased to 1.07 atm?

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