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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.

Respuesta :

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

Explanation :

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]N_2[/tex] gas = 10 atm

V = Volume of gas = 20 L

n = number of moles [tex]N_2[/tex] = 2.5 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]N_2[/tex] gas = ?

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

[tex]10atm\times 20L=2.5mole\times (0.0821L.atm/mol.K)\times T[/tex]

[tex]T=974.4K[/tex]

The temperature of the mixture will be same as the temperature of [tex]N_2[/tex] gas.

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

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