At what temperature celsius will 19.4g of molecular oxygen, o2, exert a pressure of 1820mm hg in a 5.12th cylinder?

T = - 28.2 °C

Solution:

Data Given;

P = 1820 mmHg = 2.39 atm

V = 5.12 L

m = 19.4 g

R = 0.0821 atm. L. mol⁻¹. K⁻¹

Calculating moles,

n = 19.4 g : 32 g. mol⁻¹

n = 0.606 mol

Assuming that the oxygen gas is acting perfectly, then according to Ideal Gas Equation,

P V = n R T

Solving for T,

T = P V / n R

Putting values,

T = (2.39 atm * 5.12 L) : (0.606 mol * 0.0821 atm. L. mol⁻¹. K⁻¹)

T = 244.7 K

Or,

T = - 28.2 °C