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Gas laws problem
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:
PV=nRT
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:
\frac{2.44e23+3.0e23}{6.02e23}=\sim 0.90 mol
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:
620x=0.90*8.31*273
x\sim 3.29_L
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..
3.29x=0.41*8.31*273
x\sim 282.7_{kPa}
so now by law of partial pressures 620-282.7=337.3 kPa
so my answer is:
H_2 = 282.7_{kPa}
N_2 = 337.3_{kPa}
i got it wrong, i think I am overcomplicating things, could some1 help me out?
thnx
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

so basically the way i though of solving this was to use the ideal gas law being:
PV=nRT
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:
\frac{2.44e23+3.0e23}{6.02e23}=\sim 0.90 mol
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:
620x=0.90*8.31*273
x\sim 3.29_L
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..
3.29x=0.41*8.31*273
x\sim 282.7_{kPa}
so now by law of partial pressures 620-282.7=337.3 kPa
so my answer is:
H_2 = 282.7_{kPa}
N_2 = 337.3_{kPa}
i got it wrong, i think I am overcomplicating things, could some1 help me out?
thnx