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# Alculate the Total Pressure in a Mixture of 8 G of Dioxygen and 4 G of Dihydrogen Confined in a Vessel of 1 Dm3 at 27°C. R = 0.083 Bar Dm3 K–1 Mol–1. - Chemistry

#### Question

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K1 mol1.

#### Solution 1

Given,

Mass of dioxygen (O2) = 8 g

Thus, number of moles of O_2 = 8/32 = 0.25 mole

Mass of dihydrogen (H2) = 4 g

Thus, number of moles of H_2 = 4/2 = 2 "mole"

Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole

Given,

V = 1 dm3

n = 2.25 mol

R = 0.083 bar dm3 K–1 mol–1

T = 27°C = 300 K

Total pressure (p) can be calculated as:

pV = nRT

=> p = (nRT)/V

= (225xx0.083 xx 300)/1

= 56.025 bar

Hence, the total pressure of the mixture is 56.025 bar.

#### Solution 2

Molar mass of  O_2 = 32 g mol^(-1)  :. 8gO_2  = 8/32 mol = 0.25 mol

Molar mass of H_2 =  2 g  mol^(-1)    :. 4gH_2 = 4/2 = 2 mol

:. Total number of moles(n) = 2 + 0.25 =2.25

V = 1 dm^3, T = 27^@C = 300  K, R = 0.083 "bar" dm^3 K^(-1) mol^(-1)

RV = nRT

or

P = (nRT)/V = ((2.25 mol) (0.083 "bar" dm^3 K^(-1)mol^(-1))(300 K))/(1 dm^3)

= 56.025 bar

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