Question

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

Answer 1

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

Answer 2

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

Step-by-step explanation:

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

`p_T=\sum_(i=1)^n(\chi_(i)* p_i)`

where,

`p_T` = total pressure of the mixture

`\chi_(i)` = mole fraction of i-th species

`p_i` = 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:

`p_T=[(0.5* 1.7)+(0.23* 1.1)+(0.12* 0.7)+(0.10* 0.5)+(0.05* 0.2)]`

`p_T=1.247atm`

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