#### Question

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^{3} at 27°C. R = 0.083 bar dm^{3} K^{–}^{1} mol^{–}^{1.}

#### Solution 1

Given,

Mass of dioxygen (O_{2}) = 8 g

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

Mass of dihydrogen (H_{2}) = 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 dm^{3}

*n* = 2.25 mol

R = 0.083 bar dm^{3} K^{–1} mol^{–1}

*T* = 27°C = 300 K

Total pressure (*p*) can be calculated as:

*pV* =* n*R*T*

*`=> 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