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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. - CBSE (Science) Class 11 - Chemistry

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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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Solution 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. Concept: Ideal Gas Equation - Density and Molar Mass of a Gaseous Substance.
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