2)

If 5.0 moles of O2 and 3.0 moles of N2 are placed in a 30.0 L tank at a temperature of 25°

C, what will the pressure of the resulting mixture of gases be?

Total pressure = 6.55 atm

Explanation:

Given dа ta:

Number of moles of oxygen = 5.0 mol

Number of moles of nitrogen = 3.0 mol

Volume of tank = 30.0 L

Temperature = 25°C (25+273 = 298 K)

Total pressure of mixture = ?

Solution:

Pressure of oxygen:

PV = nRT

P = nRT/V

P = 5.0 mol * 0.0821 atm. L/mol. K * 298 k / 30.0 L

P = 122.33 atm / 30.0

P = 4.1 atm

Pressure of nitrogen:

PV = nRT

P = nRT/V

P = 3.0 mol * 0.0821 atm. L/mol. K * 298 k / 30.0 L

P = 73.4 atm / 30.0

P = 2.45 atm

Total pressure = Partial pressure of oxygen + partial pressure of nitrogen

Total pressure = 4.1 atm + 2.45 atm

Total pressure = 6.55 atm