Question

A gaseous mixture is composed of 1.5 moles of He, 2.5 moles of N2 and an unknown number of moles of He. The mixture is contained within a 20-liter vessel. If the partial pressure of the N2 ​​ is 10 atm, what is the temperature of the mixture in degrees Kelvin, ˚K? R= 0.0821 L x atm/mol x ˚K.

Answer 1

Answer : The temperature of the mixture will be, 974.4 K

Explanation :

Using ideal gas equation:

`PV=nRT`

where,

P = Pressure of
`N_2` gas = 10 atm

V = Volume of gas = 20 L

n = number of moles
`N_2` = 2.5 mole

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`N_2` gas = ?

Now put all the given values in above equation, we get:

`10atm* 20L=2.5mole* (0.0821L.atm/mol.K)* T`

`T=974.4K`

The temperature of the mixture will be same as the temperature of
`N_2` gas.

Therefore, the temperature of the mixture will be, 974.4 K