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What is the total pressure of a mixture that contains 50% nitrogen at 1.7 atm, 23% oxygen at 1.1 atm, 12% argon at 0.7atm, 10% methane at 0.5 atm, and 5% water vapor at 0.2 atm?

A. 0.13 atm
B. 1.247 atm
C. 0.85 atm
D. 4.2 atm

Respuesta :

I think so 4.2, we get the answer when we add together the pressure of each of the gases.

Answer: The total pressure of a mixture is 1.247 atm.

Explanation:

To calculate the total pressure of the mixture of the gases, we use the equation given by Raoult's law, which is:

[tex]p_T=\sum_{i=1}^n(\chi_{i}\times p_i)[/tex]

where,

[tex]p_T[/tex] = total pressure of the mixture

[tex]\chi_{i}[/tex] = mole fraction of i-th species

[tex]p_i[/tex] = partial pressure of i-th species

We are given:

  • For nitrogen:

Mole fraction of nitrogen = 0.5

Partial pressure of nitrogen = 1.7 atm

  • For oxygen:

Mole fraction of oxygen = 0.23

Partial pressure of oxygen = 1.1 atm

  • For argon:

Mole fraction of argon = 0.12

Partial pressure of argon = 0.7 atm

  • For methane:

Mole fraction of methane = 0.10

Partial pressure of methane = 0.5 atm

  • For water vapor:

Mole fraction of water vapor = 0.05

Partial pressure of water vapor = 0.2 atm

Putting values in above equation, we get:

[tex]p_T=[(0.5\times 1.7)+(0.23\times 1.1)+(0.12\times 0.7)+(0.10\times 0.5)+(0.05\times 0.2)][/tex]

[tex]p_T=1.247atm[/tex]

Hence, the total pressure of a mixture is 1.247 atm.

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