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Why does nitrogen gas' pressure not change after an addition of helium 
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#1
Apr1014, 08:58 PM

P: 6

a 9 L vessel contains 3 moles of helium and 3 moles of nitrogen at a pressure of 10 atm. Maintaining constant temperature, an additional 2 moles of helium are added. Assuming gases behave ideally, what are the partial pressures of nitrogen and helium at the end?
Initially there's: 3mol He/(3mol + 3mol) * 10 atm = 5 atm pressure of He And: 3/6 * 10atm = 5 atm pressure of N_{2} as well now that 2 moles of He are added to the vessel, why wouldn't N_{2}'s partial pressure increase? The way I see it, there are now more atoms, so more of them would knock on the walls, therefore pressure would increase overall, and since there are now more atoms, space between atoms is tightened, so ... Anyway, overall pressure is P_{1}/n_{1} = P_{2}/n_{2} (10 atm/9 mol) = (P_{2}/11 mol); P_{2} = 12.2 atm new total pressure Partial P_{He} * P_{total} = 5/11 * 12.2 atm = 5.545 atm He That's for He gas, but why shouldn't partial pressure change for N_{2} gas? 


#2
Apr1114, 02:02 AM

HW Helper
P: 2,263

There are the same amount of nitrogen molecules so the nitrogen partial pressure does not change. Your total pressure is wrong. Where did 9 and 11 come from?



#3
Apr1214, 05:04 PM

P: 380

This comes from the assumption of ideal gas behavior. The partial pressure of an ideal gas is the pressure of the same amount of the pure gas at the same temperature/volume as in the mixture. Ideal gases are assumed to have zero intermolecular interactions and particle volume, this greatly simplifies a lot of the mathematics and is a very good approximation to small (mon and diatomic) gases at fairly low pressures. The greater the pressure or the larger the gas the greater the deviation from ideality which necessitates the use of "messy" virial or Van der Waals equation to account for the intermolecular forces and nonnegligible particle volume.
I'd recommend getting comfortable with the assumptions underlying the concept of an ideal gas. 


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