Some problems

1) Air may be considered as a mixture of 78.1% nitrogen, 21.0% oxygen and 0.9% argon by volume. Calculate the partial pressure of each gas in air at room temperature.

Do I first find the mole of the gas, let's say, nitrogen.

78.1 / 14.01 = 5.57 mole

And then use PV = nRT to solve the question?

But then I do not know the volume of the gases.
And I'm not sure whether what I calculated for the mole of nitrogen is correct or not.
Room temperature is 273 + 24 = 297K right?

2)
a) Predict the volume, which 96g of oxygen gas would occupy at 100 degree celcius and 5065 kpa.
(I can do this part and the answer is 1.83 something, I've problem on the next part.)
The experimentally measured value for this volume is 1.81 liters. Explain why this differs from the volume predicted in a).

Thanks for helping.

Reply 1

Kyalimers

Ok so for Q1:

The volumes are relative percentage wise are they not!!!
And yes you are correct: the molar value for Nitrogen is wrong.

Q2:

Think about what can go wrong during the Pv=nRT experiment. HINT: Is the temperature always 25 deg C. How accurate is the data used etc...

Reply 2

Civ-217

sohanshah

Can you show me how to calculate the partial pressure for nitrogren? Then I'll understand and apply it on the other. Because I really don't know how to do this question, even after you giving hint.

2)Can I say PV=nRT is an ideal gas equation whereas we deal with real gases in reality? I don't understand about the temperature because I thought the temperature in this question is 100 degree celcius.

Thanks for helping.

Reply 3

Kyalimers

For 2) you hit the nail bang on. The eq. is ideal!!! but we live in reality. For the temp and pressure, what I was saying is in reality the temp and pressure will change constantly - even small fluctuations will give different answers!!!

For 1) they have to have given you more info surely....

Reply 4

Kyalimers

BTW what year are you: because if you are in year 13 you should have covered partial pressure constant equations and the mole fraction.

Reply 5

Kyalimers

so the mole fraction of Nitrogen would be 0.78, for oxygen it would be 0.21 and for Argon it would be 0.9

Reply 6

Sinuhe

1. If we assume the percentages are given for volume/volume ratios (and not mass/mass) - this should probably be stated explicitly however, then we can note that for a perfect gas, n (the amount, in moles) is proportional to the volume at constant pressure and temperature, so this is ALSO a mole/mole ratio, i.e. the mole fraction (usually denoted X). So the problem is trivial. By definition, for species i, the partial pressure is given by:

p_{i} = X_{i} \cdot p_{\mathrm{total}}

So, since you know that atmospheric pressure is 1 atm,

p_{N_2}=0.781 \cdot 1~\mathrm{atm}=79.134~\mathrm{kPa}

, and so on.
You actually don't need to use the ideal gas law at all.

Also, regarding your calculation:
- no, the amount of nitrogen is NOT 5.57 mol; you don't know how much nitrogen is in air, surely. Also, the molar mass of nitrogen is not 14 g/mol but 28 g/mol (the gas is dinitrogen, not atomic nitrogen).

- room temperature in chemistry is generally defined to be 25 degrees Celsius, i.e. 298 K.

As for 2., you have the right idea, real gases generally don't obey the perfect gas law over any significant range of pressures/temperatures. You can define the compression factor, often denoted Z, to measure the deviation from ideality, such that

Z=\frac{V_m}{V_m^{\circ}}

where

V_m

is the molar volume and

V_m^\circ

is the molar volume of a perfect gas at the same pressure and temperature.
There are three regimes you need to consider:
- Z=1 - ideal gas behaviour (when there are no forces, or when all forces are cancelled out (at the Boyle temperature for example))
- Z<1 - attractions dominate, the gas occupies a smaller volume than an ideal gas would
- Z>1 - repulsions dominate, bigger volume

At very large separations, forces between gas molecules will be ~ 0 and hence Z=1. This occurs at low pressures.
At small separations, attractions (such as van der Waals forces) dominate, and molecules are attracted together more, Z<1.
Finally, at very small separations, the electron clouds of individual molecules begin to overlap, leading to very strong repulsions, repelling the molecules from each other. Then Z>1.

50 atm is a medium pressure, so you can expect oxygen to be in the second regime, with Z<1. So the volume (as the molar volume) will be less than calculated from the perfect gas law. This is indeed what you have been told in the question.

Reply 7

Kyalimers

Was that necessary??? You have just confused him more!!!

Reply 8

Sinuhe

sohanshah

Was that necessary??? You have just confused him more!!!

Which part did you find confusing? I can rephrase it if you find it's badly worded - but it seems very straightforward to me ... :confused:

Also, I don't think it's right to say that 'For the temp and pressure, what I was saying is in reality the temp and pressure will change constantly - even small fluctuations will give different answers!!!' - it is not the case that we cannot fix the temperature or pressure precisely, because we can. In other words, it's not an experimental failure that you get a different answer, but a fundamental deviation from predicted behaviour that can be explained by considering molecular interactions.

Reply 9

Kyalimers

Precisely... but the accuracies of the temperature and pressure will always change. So i called them fluctuations. You are deeply and technically correct but for such a simple question I believe youa re over complicating it.

Reply 10

Kyle_S-C

Yeah, but the OP doesn't necessarily have to include that in their answer, nor read the post at all. It's kinda interesting to be given the actual explanations behind things, I'd have probably have been less helpful and pointed out the Van der Waals' equation and even put in a space to separate out my words.