Question

A mixture of He He , N 2 N2 , and Ar Ar has a pressure of 22.1 22.1 atm at 28.0 28.0 °C. If the partial pressure of He He is 2475 2475 torr and that of Ar Ar is 1049 1049 mm Hg, what is the partial pressure of N 2 N2 ?

Answer 1

Answer:

17.47 atm

Explanation:

We need to do some conversions of units to determine this answer

Mixture of gases: He, N₂ Ar

Total pressure: 22.1 atm

Sum of partial pressures of each gas from the mixture = Total pressure

Partial pressure He + Partial pressure N₂ + Partial pressure Ar = 22.1 atm

Partial pressure N₂ = 22.1 atm - Partial pressure He - Partial pressure Ar

We convert partial pressures Ar and He to atm

2475 Torr . 1atm / 760 Torr = 3.25 atm → Partial pressure He

1049 mmHg . 1 atm / 760 mmHg = 1.38 atm → Partial pressure Ar

We replace data:

Partial Pressure N₂ = 22.1 atm - 3.25 atm - 1.38 atm → 17.47 atm

Answer 2

The partial pressure of N2 when the partial pressure of He is 24751 should be 7.47 atm.

Calculation of the partial pressure:

Since the Mixture of gases: He, N₂ Ar

And, Total pressure: 22.1 atm

So,

Sum of partial pressures of each gas from the mixture = Total pressure

Like wise

Partial pressure He + Partial pressure N₂ + Partial pressure Ar = 22.1 atm

Here,

Partial pressure N₂ should be

= 22.1 atm - Partial pressure He - Partial pressure Ar

Now we have to convert partial pressures Ar and He to atm

So,

2475 Torr . 1atm / 760 Torr = 3.25 atm → Partial pressure He

1049 mmHg . 1 atm / 760 mmHg = 1.38 atm → Partial pressure Ar

So,

Partial Pressure N₂ = 22.1 atm - 3.25 atm - 1.38 atm

= 17.47 atm

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