Question

a mixture of he, n₂, and ar has a pressure of 17.2 atm at 28.0°c. if the partial pressure of he is 2299 torr and that of ar is 2237 mm hg, what is the partial pressure of n₂?

pn₂ =

atm

Explanation:

Step1: Convert units

Convert the partial pressures of \(He\) and \(Ar\) to \(atm\).
For \(He\): \(P_{He}=\frac{2299\ torr}{760\ torr/atm}\approx3.025\ atm\)
For \(Ar\): \(P_{Ar}=\frac{2237\ mmHg}{760\ mmHg/atm}\approx2.943\ atm\)

Step2: Use Dalton's law of partial pressures

Dalton's law states \(P_{total}=P_{He}+P_{Ar}+P_{N_{2}}\).
Rearrange for \(P_{N_{2}}\): \(P_{N_{2}}=P_{total}-P_{He}-P_{Ar}\)
Substitute values: \(P_{N_{2}} = 17.2\ atm-3.025\ atm - 2.943\ atm\)
\(P_{N_{2}}=11.232\ atm\approx11.2\ atm\)

Answer:

\(11.2\ atm\)