Question

A mixture of N2, O2 and He have mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pressure of the mixture is 3.9 atm?

Answer 1

Answer:

Partia pressure N₂ → 0.975 atm

Explanation:

Let's analyse the moles fractions:

N₂ → 0.25

O₂ → 0.65

He → 0.1

Partial pressure / Total pressure = Mole fraction

Partial pressure N₂ / 3.9 atm = Mole fraction N₂

Partial pressure N₂ / 3.9 atm = 0.25

Partial pressure N₂ = 3.9 atm . 0.25 → 0.975 atm

Answer 2

The pressure of nitrogen in the mixture has been 0.975 atm.

Partial pressure has been defined as the pressure exerted by the gas molecules in the mixture. The partial pressure ($P_A$)  has been expressed as:

$P_A=X_A\;\times\;P$

Where,  Mole fraction of the element A, $X_{N_2}=0.25\;\rm atm$

The total pressure of the mixture, $P=3.9\;\rm atm$

Substituting the values, the pressure of nitrogen ($P_{\rm N_2}$),

$P_{\rm N_2}=0.25\;\times\;3.9\;\rm atm\\ \textit P_{N_2}=0.975\;atm$

The pressure of nitrogen in the mixture has been 0.975 atm.

For more information about mole fraction, refer to the link:

https://brainly.com/question/2060778