A 642 mL sample of oxygen gas at 23.5°C and 795 mm Hg, is heated to 31.7°C and the volume of the gas expands to 957 mL. What is the new pressure in atm?

Let's assume that O₂ is an ideal gas.

We can use combined gas law,

PV/T = k (constant)

Where, P is the pressure of the gas, V is volume of the gas and T is the temperature of the gas in Kelvin.

For two situations, we can use that as,

P₁V₁/T₁ = P₂V₂/T₂

P₁ = 795 mm Hg

V₁ = 642 mL

T₁ = (273 + 23.5) K = 296.5 K

P₂ = ?

V₂ = 957 mL

T₂ = (273 + 31.7) K = 304.7 K

By applying the formula,

795 mm Hg x 642 mL / 296.5 K = P₂ x 957 mL / 304.7 K

P₂ = 548.07 mm Hg

P₂ = 548 mm Hg

760 mmHg = 1 atm

548 mm Hg = 1 atm x (548 mmHg / 760 mmHg) = 0.721 atm

Pressure of gas = 548 mm Hg = 0.721 atm