Question

A mixture of He, N2 , and Ar has a pressure of 23.5 atm at 28.0 °C. If the partial pressure of He is 1575 torr and that of Ar is 791 mm Hg, what is the partial pressure of N2 ?

Answer 1

To calculate the partial pressure of N2, the partial pressures of He and Ar are subtracted from the total pressure after converting their values to atm. The resulting partial pressure of N2 is 20.387 atm.

Step-by-step explanation:

To find the partial pressure of N2, we will use Dalton's law of partial pressures. According to Dalton's law, the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. The problem states that the total pressure of a mixture of He, N2, and Ar is 23.5 atm. We are given the partial pressures of He and Ar in different units (torr and mm Hg), which are equivalent, so we need to convert them to atm.
1 atm = 760 torr = 760 mm Hg.

Partial pressure of He in atm: 1575 torr / 760 torr/atm = 2.072 atm
Partial pressure of Ar in atm: 791 mm Hg / 760 mm Hg/atm = 1.041 atm

Now, we can find the partial pressure of N2 by subtracting the partial pressures of He and Ar from the total pressure:

PTotal = PHe + PN2 + PAr
PN2 = PTotal - PHe - PAr
PN2 = 23.5 atm - 2.072 atm - 1.041 atm
PN2 = 20.387 atm

Therefore, the partial pressure of N2 in the mixture is 20.387 atm.