Question

56 g of N and 96 g of O are mixed in 2 litrs container at 300 k. calculate the partial pressure of each gas and total pressure of the mixeure​

Answer 1

Answer:

Given Mass of N = 56 g

Given Mass of O = 96 g

Moles of N = 56/28 = 2 moles

Moles of O = 96/32 = 3 moles

Mole Fraction of N = 2/ ( 2 + 3) = 2/5

Mole Fraction of O = 3/5

Partial pressure of N, P = KH (X)

P = KH ( 2/5)

KH for N2 = 6.51 × 10^7 mm

P(N) = 6.51 × 2 × 10^7 /5 = 2.604 mm

Partial Pressure of O, P = KH ( 3/5)

KH for O2 = 3.30 × 10^7 mm

P(O) = 3.30× 3 × 10^7 mm/ 5 = 1.98 × 10^7 mm

Total Pressure of the mixture = P(N) + P(O) = ( 2.60 + 1.98 ) × 10^7 mm

Total Pressure, P(T ) = 4.58 × 10^7 mm.