** 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:

P_{1} = 722 mmHg

${\mathbf{P}}_{\mathbf{1}}\mathbf{=}\mathbf{722}\mathbf{}\overline{)\mathbf{mmHg}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{atm}}{\mathbf{760}\mathbf{}\overline{)\mathbf{mmHg}}}$

P_{2} = **0.95 atm**

P_{2} = **1.07 atm**

T_{1} = 22°C +273.15 = **295.15 K**

T_{2} =** ??**

We can rearrange the equation to solve for T_{2}:

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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