Calculate the total pressure in a mixture of `8 g` of oxygen and `4 g` hydrogen confined in a vessel of `1 dm^(3)` at `27^(@)C`. `(R = 0.083 "bar" dm^(3) K^(-1) mol^(-1))`

Answer

65.025 bar

Explanation

Given, mass of dioxygen `(O_(2))` = 8 g
Thus, numbe of moles of `O_(2)=8/32=0.25`mole
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`=300K
Total pressure (p) can be calculate as :
pV= nRT
`rArr=(nRT)/V`
`=(225xx0.083xx300)/1`
=56.025 bar
Hence, the total pressure of the mixture is 56.025 bar.