2.44e23 molecules of Hydrogen and 3.0 molecules of Nitrogen are together exerting a pressure of 620. kPa. what is the partial pressure of each gas?

**Ans_2 = 278 kPa, N_2 = 342 kPa**

so basically the way i though of solving this was to use the ideal gas law being:

[tex]PV=nRT[/tex]

i kept the temperature constant at 273 K. then i found the total amount of moles by adding the # of molecules and dividing by 6.02e23:

[tex]\frac{2.44e23+3.0e23}{6.02e23}=\sim 0.90 mol[/tex]

so then i asked myself: at constant temperature, what would be the total volume these gases would occupy at 620 kPa and 0.90 mol. so i solved:

[tex]620x=0.90*8.31*273[/tex]

[tex]x\sim 3.29_L[/tex]

so now i took hydrogen, at constant temperature, occuping a volume of 3.29 L, how much pressure would it exert? 2.44e23 is 0.41 mol so..

[tex]3.29x=0.41*8.31*273[/tex]

[tex]x\sim 282.7_{kPa}[/tex]

so now by law of partial pressures 620-282.7=337.3 kPa

so my answer is:

[tex]H_2 = 282.7_{kPa}[/tex]

[tex]N_2 = 337.3_{kPa}[/tex]

i got it wrong, i think im overcomplicating things, could some1 help me out?

thnx