Question

A mixture of nitrogen (N2) and helium has a volume of 290 mL at 26 ∘C and a total pressure of 745 mmHg .

(A.) If the partial pressure of helium is 35 mmHg , what is the partial pressure of the nitrogen?

(B.) At this temperature and at the partial pressure in Part A, how many moles of nitrogen gas are in this container assuming ideal conditions?

Answer 1

(A) Given total pressure P=745 mmHg, and P_He=35 mmHg, P_N2=?

=> The total pressure of an ideal gas mixture is the sum of the partial pressuresof each individual gas in the mixture.

P= P_He+P_N2

745 mmHg=35 mmHg+P_N2

P_N2=710 mmHg

(B) From ideal gas law PV=nRT,

Given V=290 ml=0.290 L, T=26 °C=26+273=299 K, P_N2=710 mmHg=0.934 atm, R=8.2*10-2 L atm K-1 mol-1

n=?

n=PV/RT=(0.934 atm*0.290 L)/(8.2*10-2 L atm K-1 mol-1*299 K)=0.011 mol N2

Moles of nitrogen gas=0.011 mol.