# 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:

$\overline{)\frac{{\mathbf{P}}_{{\mathbf{1}}}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{=}}\frac{{\mathbf{P}}_{{\mathbf{2}}}}{{\mathbf{T}}_{\mathbf{2}}}}$

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

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:

88% (211 ratings) ###### 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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