A sample of 2.00 moles of oxygen at 45°C and 98.6 kPa occupies what volume?

V = 53.83 L

Explanation:

Given data:

Number of moles = 2.00 mol

Temperature = 45°C (45+273 = 318 K)

Pressure = 98.6 KPa (98.6/101 = 0.97 atm)

Volume of gas = ?

Solution:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

0.97 atm × V = 2.00 mol × 0.0821 atm.L/mol. K × 318 K

V = 52.22 atm.L/0.97 atm

V = 53.83 L

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