[sfunds]

Calculate the partial pressure N2 and H2 in a mixture of two moles of n2 and two moles of h2 at stp in a open vessel.
Sol.  Pn2 = number of moles of n2  X RT / V
  ph2 = number of moles of h2 X RT/V
Mole fraction of N2 = 2/(2+2) = 0.5
Mole fraction of H2 = 0.5
 
But P = RT/V for 1 mole V = 22.4 lt
for 4 Mol 4 X22.4 lt
R = .0821 lit-atmK-1mol-1 (-1 denotes power)
P = .0821 X 273/4 X 22.4  = .2501
Pn2 = ph2= .1251atm
This is how the problem is solved in chemistry XI matriculation board.(TN)
My doubt is PV = nRT
so while calculating Pressure why n is not taken into consideration
ie P = nRT/V = 4* .0821*273/4*22.4
can somebody clearly explain when n should be in the formula

 Post link to some nice site explaining Mole concept and partial pressure

[ultrashogun]

for 4 Mol 4 X22.4 lt

This is where you put n into the formula. Its a confusing way, but they calculated the pressure for 1 mol, then multiplied the whole thing by 4. If you neglect a member of the formula consider it as 1.

There always has to be n in the equation.