[student15]

Hey guys, I'm new here I was wondering if you could help me with a problem I'm stuck on.
I've gotten as far as the total mole but that is.
Problem:
A mixture of argon and nitrogen is introduced inside a 234 dm^3 vessel. At 298.5 the pressure of the mixture is measured to be 7.55 kPa. And the total mass of the gas is 33.28g. Determine the composition of the argon and nitrogen mixture in mole fractions.

Thank you. I will try to work out in the mean time.

[Cooper]

Hey guys, I'm new here I was wondering if you could help me with a problem I'm stuck on.
I've gotten as far as the total mole but that is.
Problem:
A mixture of argon and nitrogen is introduced inside a 234 dm^3 vessel. At 298.5 the pressure of the mixture is measured to be 7.55 kPa. And the total mass of the gas is 33.28g. Determine the composition of the argon and nitrogen mixture in mole fractions.

Thank you. I will try to work out in the mean time.

$$m_{TOT}=n_{N_2}M_{N_2}+n_{Ar}M_{Ar}\\n_{TOT}=n_{N_2}+n_{Ar}$$
When you have two equations you can solve for one variable in one of them, say n_N2 in the second equation, and substitute it into the first equation. This will enable you to solve for the quantity n_Ar and subsequently n_N2.

[student15]

Hey guys, I'm new here I was wondering if you could help me with a problem I'm stuck on.
I've gotten as far as the total mole but that is.
Problem:
A mixture of argon and nitrogen is introduced inside a 234 dm^3 vessel. At 298.5 the pressure of the mixture is measured to be 7.55 kPa. And the total mass of the gas is 33.28g. Determine the composition of the argon and nitrogen mixture in mole fractions.

Thank you. I will try to work out in the mean time.

$$m_{TOT}=n_{N_2}M_{N_2}+n_{Ar}M_{Ar}\\n_{TOT}=n_{N_2}+n_{Ar}$$
When you have two equations you can solve for one variable in one of them, say n_N2 in the second equation, and substitute it into the first equation. This will enable you to solve for the quantity n_Ar and subsequently n_N2.

So if i have n(total) which is .712 mol. Given the fact that the mixture weights 33.28 g, I get 33.28 g / .712 mol to get 46.7416 g/mol. Is it ok to assume that that is the molecular weight for the mix?

[Cooper]

Is it ok to assume that that is the molecular weight for the mix?

Well yes I guess it would be the molecular weight of the mixture but you don't need that quantity to solve this problem.

[student15]

Is it ok to assume that that is the molecular weight for the mix?

Well yes I guess it would be the molecular weight of the mixture but you don't need that quantity to solve this problem.

ok I tried to work it out but it does not seem to work. x=mol of N₂ y=mol of Ar
33.28 g = (x)(28.014g/mol)+(y)(39.948g/mol)
gives x=(33.28 g - (y)(39.948g/moll))/28.017g/mol
substituting in gives 33.28 g = (33.28 g - (y)(39.948g/moll))/28.017g/mol)*28.017g/mol + (y)(39.948)
After everything cancels out it just gives 33.28=33.28 .
So am I doing something wrong?

[Cooper]

substituting in gives 33.28 g = (33.28 g - (y)(39.948g/moll))/28.017g/mol)*28.017g/mol + (y)(39.948)
After everything cancels out it just gives 33.28=33.28 .
So am I doing something wrong?

Yes, if you think about it solving for a variable in an equation and then substituting that back into the same equation should just tell you 1=1. The trick here is that you have to use two equations.

What do we know?
$$m_{TOT}=n_{N_2}M_{N_2}+n_{Ar}M_{Ar}\\n_{TOT}=n_{N_2}+n_{Ar}$$

Since it's easier than doing it with the first equation, let's solve for n_N2 with the second equation...

$$n_{N_2}=n_{TOT}-n_{Ar}$$

Now if you put this substitution for n_N2 into the first equation (for m_TOT) you will be able to solve for n_Ar.

[student15]

substituting in gives 33.28 g = (33.28 g - (y)(39.948g/moll))/28.017g/mol)*28.017g/mol + (y)(39.948)
After everything cancels out it just gives 33.28=33.28 .
So am I doing something wrong?

Yes, if you think about it solving for a variable in an equation and then substituting that back into the same equation should just tell you 1=1. The trick here is that you have to use two equations.

What do we know?
$$m_{TOT}=n_{N_2}M_{N_2}+n_{Ar}M_{Ar}\\n_{TOT}=n_{N_2}+n_{Ar}$$

Since it's easier than doing it with the first equation, let's solve for n_N2 with the second equation...

$$n_{TOT}-n_{Ar}$$

Now if you put this substitution for n_N2 into the first equation (for m_TOT) you will be able to solve for n_Ar.

$$m_{TOT}=n_{TOT}-n_{Ar}M_{N_2}+n_{Ar}M_{Ar}$$
$$33.28 g = (.712g/mol-x)(28)+x(40)$$
$$33.28 g = 19.936-28x+40x$$
$$13.344 = 12x$$
$$x=1.112$$
Which should not be possible since there is only .712 mol total.

[student15]

PV=nRT
PV/RT=n
(7.55 kPa)(234 L)/(8.314 J/molK)(298.5K) =n
n_total=.712 mol
n=m/M
M=m/n
Assuming we can use the molecular weight as a mixture of both gasses
M=(33.28 g)/(.712 mol)
M=46.7416 g/mol
Now this is impossible since combination of Argon and Nitrogen could never go above 39.948g/mol which is the MW of argon

[Borek]

There is something wrong with the data given. Assuming pure Ar, mass of the 0.712 moles of the gas would be 28.4 g. Assuming pure nitrogen - 19.9 g. Mass given is not between those two values, so it is impossible.

[student15]

There is something wrong with the data given. Assuming pure Ar, mass of the 0.712 moles of the gas would be 28.4 g. Assuming pure nitrogen - 19.9 g. Mass given is not between those two values, so it is impossible.

I got an email from my teacher telling me the total mass given was a typo. Thank you.