Molar Ratio of Two Ideal Gases

In summary, Homework Equations state that the ratio of moles of gas B to moles of gas A is in terms of p1, p2, T1 and T2.
  • #1
enokoner
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Homework Statement



A mixture of two gases, A and B, exists at pressure p1, volume V, and temperature T1. Gas A is subsequently removed from the mixture in a constant-volume process. The remaining gas B is found to have a pressure p2, volume V, and temperature T2. Express the ratio of the number of moles of gas B to the number of moles of gas A in the terms of p1, p2, T1 and T2.

a. [itex] \frac{p_2 T_1}{p_1 T_2 - p_2 T_1}[/itex]

b. [itex] \frac{p_2 T_1^{2}}{T_2(p_1 T_2 - p_2T_1)}[/itex]


*Options c and d were not written because they contained specific gas constants which do not pertain to molar equations.




Homework Equations




Ideal Gas: [itex]pV = N \overline{R}T[/itex]

Dalton's Law: [itex] p = \sum p_i [/itex]


The Attempt at a Solution



Universal gas constant crosses out. Volume stays constant and also crosses out.

∴ [itex] \frac{N_B}{N_A}= \frac{p_B T_A}{T_B p_A} [/itex]


Relating pA and pB to p1

[itex] p_1= p_A + p_B [/itex]


and to p2

[itex] p_2 = p_B [/itex]


Also, [itex] T_A = T_1 \ T_B = T_2 [/itex]


∴ [itex] \frac{N_B}{N_A}= \frac{p_2 T_1}{T_2(p_2 - p_1)}[/itex]


This is not an option. I have a feeling its because I assumed [itex] p_2 = p_B [/itex]. That assumption doesn't feel right. I don't know how else to relate these two. Thank you for considering this.
 
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  • #2
enokoner said:
[itex] \frac{N_B}{N_A}= \frac{p_B T_A}{T_B p_A} [/itex]

I am not sure I get it.

I have a feeling you can be a victim of your own choice of confusing indices.
 
  • #3
Borek said:
I am not sure I get it.

I have a feeling you can be a victim of your own choice of confusing indices.

I agree that the indices are out of control. But I double checked. Sorry for the step-skipping.

[itex]

N = \frac{pV}{\overline{R}T} \\

∴ \frac{N_B}{N_A} = \frac{\frac{p_BV}{\overline{R}T_B}}{\frac{p_A V}{\overline{R}T_A}} \\

=\frac{p_B T_A}{p_A T_B}


[/itex]
 
  • #4
enokoner said:
[itex] p_1= p_A + p_B [/itex]

Aren't pA and pB measured at different temperatures?

I got 'a', starting from

[tex]p_1V=(N_A+N_B)RT_1[/tex]

[tex]p_2V=N_BRT_2[/tex]
 
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  • #5
OP Second Attempt by solving in reverse.

I got the answer but I don't know how. The book states that the answer is B. I got it by solving the problem in reverse. I used a common equation that relates pressure and temperature wrt Ideal gases. The problem is I don't understand why I'm supposed to used that or why I should set certain values to each other.2.1 Relevant equationsIdeal Gas: [itex]pV = N \overline{R}T[/itex]

Dalton's Law: [itex] p = \sum p_i [/itex]

Assuming System is isentropic: [itex] \frac{T_2}{T_1} = (\frac{p_2}{p_1})^{\frac{k-1}{k}} [/itex]


3. Second attempt at a solution

Universal gas constant crosses out. Volume stays constant and also crosses out.

∴ [itex] \frac{N_B}{N_A}= \frac{p_B T_A}{T_B p_A} [/itex]Relating pA and pB to p1:

[itex] p_1= p_A + p_B [/itex]and to p2:

Assuming (I HAVE NO BASIS FOR THIS ),

[itex] p_B = p_2 ; \ p_2 = p_1 ; \ k = 0.5 [/itex] Therefore,

[itex] \frac{T_2}{T_1} = (\frac{p_B}{p_2})^{\frac{0.5-1}{0.5}} \\

= \frac{P_2}{P_B} \\

∴ p_B = p_2 \frac{T_1}{T_2} \\

∴ p_A = p_1 - p_B = p_1 - p_2(\frac{T_1}{T_2}) [/itex]

Also, [itex] T_A = T_1 ; \ T_B = T_2 [/itex]Substituting,

[itex] \frac{N_B}{N_A}=\frac{p_2(\frac{T_1}{T_2})T_1}{T_2(p_1-p_2(\frac{T_1}{T_2})} \\= \frac{p_2 T_1^{2}}{T_2(p_1 T_2 - p_2 T_1)} [/itex]I know setting those values arbitrarily is nonsense. But I thought perhaps it could give me some insight. It has not. I'm grasping at straws. Can anyone help?
 
Last edited:
  • #6
Thank you so much.

I got your same answer using your very clear setup. (You rock man). The solution was too good. It was elegant and made sense the whole way through...

Setting both eqs. equal to V/R:

[itex]\frac{N_A}{N_B}+ \frac{N_B}{N_B} =\frac{T_2 p_1}{p_2 T_1} \\

\frac{N_B}{N_A} = \frac{1}{\frac{T_2 p_1}{p_2 T_1} -1} \\= \frac{p_2 T_1}{T_2 p_1 -p_2 T_1}
[/itex]

Thats how it should be. So I went hunting for the online errata. I can't believe I didn't do this sooner!

errata.png

Amazing Borek! I got to go over all the errata. Thanks a million.
 
Last edited:

What is the molar ratio of two ideal gases?

The molar ratio of two ideal gases is the ratio of the number of moles of one gas to the number of moles of the other gas in a given system. It is a measure of the relative quantities of the two gases in the system.

How is the molar ratio of two ideal gases calculated?

The molar ratio of two ideal gases can be calculated by dividing the number of moles of one gas by the number of moles of the other gas. This can be done using the ideal gas law, which states that the pressure, volume, and temperature of an ideal gas are related by the equation PV = nRT, where n is the number of moles of the gas.

What is the significance of the molar ratio of two ideal gases?

The molar ratio of two ideal gases is important because it can help determine the behavior of the gases in a system. It can also be used to calculate the partial pressure of each gas, which is useful in understanding the overall pressure of the system.

What is the difference between molar ratio and mole fraction?

The molar ratio is the ratio of the number of moles of one gas to the number of moles of another gas, while mole fraction is the ratio of the number of moles of one gas to the total number of moles in the system. Molar ratio is a measure of the relative quantities of two gases, while mole fraction is a measure of their relative concentration.

How does the molar ratio of two ideal gases change with temperature and pressure?

The molar ratio of two ideal gases is constant at a given temperature and pressure, according to the ideal gas law. However, as temperature and pressure change, the number of moles of each gas may also change, leading to a change in the molar ratio. This can affect the behavior of the gases in the system.

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