Question

A special gas mixture used in bacterial growth chamber contains 1% by weight of CO2 and 99% of O2.what is the partial pressure(in atm) of is each gas as total pressure of 0.977atm?
Ans. Po2 = 0.970atm, Pco2 =0.1atm​

Answer 1

according to Dalton's law of partial pressure,

partial pressure of any gas is product of mole fraction of that gas and total pressure.

so, we should first find out mole fraction of each gas.

a/c to question,

1% by weight of CO2 and 99% by weight of O2 contains in the mixture.

if we assume weight of mixture is 100g

then, weight of CO2 = 1g while weight of O2 = 99g

now, mole of CO2 = 1/44 and mole of O2 = 99/32 [ as molar weight of CO2 = 44g/mol and molar weight of O2 = 32g/mol]

so, mole fraction of CO2 = (1/44)/(1/44 + 99/32) = 0.00729

and mole fraction of O2 = 1 - 0.00729 =0.99271

then, partial pressure of CO2 = 0.00729 × 0.977 = 0.00712233 atm ≈ 0.007atm

partial pressure of O2 = 0.99271 × 0.977= 0.96987767 atm ≈ 0.970 atm

Answer 2

Answer:

Po2 = 0.970atm; Pco2 = 0.00711atm

Explanation:

44g co2----> 1 mole

1g co2------> 1/44mole ----(I)

32g o2-----> 1 mole

99g o2----> 99/32 mole ----(II)

Xco2 = 1/44 / (1/44+99/32)

= 1/44 / (8+1089)/352

= 1/44 x 352/1097

= 8/1097

= 0.00729----(I)

Xo2 = 1-0.00729

= 0.99271-----(II)

Pco2/0.977 = 0.00712

Pco2 = 0.00712---(I)

Po2/0.977 = 0.99271

Po2 = 0.967.