What Is the Specific Heat Ratio for a Helium and Nitrogen Gas Mix?

[mk9898]

Homework Statement

A vessel contains a mix of helium and nitrogen (ideal gases). The ratio of the partial pressures is known: $\frac{p_{N2}}{pHe} = 5$. The partial pressure is the pressure that a single component alone would exert.

Homework Equations

What is the value of $\kappa = c_p/c_V$ for the mixed gas

The Attempt at a Solution

$f_{N_2} = 5$ and $f_{He} = 3$. $\kappa = 1 + \frac{2}{f}$

Of the mole mixture, N_2 pertains 5 times more moles than He. That means:

$\kappa = \frac{5}{6}(1 + \frac{2}{5}) + \frac{1}{6}(1 + \frac{2}{3})$

 

[kuruman]

If $\kappa=c_p/c_v$, shouldn't you find the weighted values of $c_v$ and $c_p$ for the mixture separately and then take the ratio?

 

[mk9898]

I'm not sure. Meaning:

$c_v = \frac{5}{6}(\frac{5*k}{2m_{N_2}}) + \frac{1}{6}(\frac{3*k}{2m_{He}})$
$c_p = \frac{5}{6}(c_v+\frac{k}{m_{N_2}}) + \frac{1}{6}(c_v+\frac{k}{m_{He}})$

Wouldn't that be redundant?

 

[kuruman]

Isn't it true that, in these units and since you will be taking ratios anyway, c_v = 3 for He and c_v=5 for N₂? Isn't it also true that c_p = 1 + c_v? Or are we talking about different things past each other?

 

[mk9898]

So then

$\kappa = \frac{\frac{5}{6}(c_v+\frac{k}{m_{N_2}}) + \frac{1}{6}(c_v+\frac{k}{m_{He}})}{\frac{5}{6}(\frac{5*k}{2m_{N_2}}) + \frac{1}{6}(\frac{3*k}{2m_{He}})}$

Right?

 

[kuruman]

Looks right, but what is $k/m$?

 

[mk9898]

Ok I made some changes to the equations once again. It is in fact $c_p = c_v + \frac{k}{m}$ and not $c_p = c_v + 1$. Given my lecture notes and textbook. Where do you get $c_p = c_v + 1$?

 

[kuruman]

I am familiar with $c_p=c_v+R$. If the specific heats are expressed in units of $R$, then $c_p=c_v+1$. That's why I asked what $k/m$ is in #6. Are we on the same page?

 

[mk9898]

Yea ok I believe yours is molar and my equation is without moles considered. k ist the Boltzmann-Constant and m is the mass of the molecules. We are on the same page. Could you give me your thoughts on the answer in #5?

 

[kuruman]

#5 looks right.

 

[mk9898]

Cool thanks. Just a side question. Do you know a website/book that goes through the derivation of your equation? I spent quite a bit of time trying to find a compete derivation but I just find tidbits here and there that just confuses me even more. I also have molar capacities in my lecture notes and they for example, that

$C_p = \frac{f+2}{2}R$

That looks much different than yours.

 

[kuruman]

That looks much different than yours.

Not really. For a monatomic gas like helium $f=3$ so that $C_v=(3/2)R$ and $C_p=(5/2)$.

Do you know a website/book that goes through the derivation of your equation?

The derivation is simple.
Start with the First Law for an isobaric (constant pressure) process
$\Delta U = Q-W$ where $W$ = work done by the gas.
Substitute $\Delta U = C_v\Delta T$.
By the definition of $C_p$, $Q=C_p \Delta T$
The work done by the gas at constant pressure is $W=p\Delta V$ and from the ideal gas law when the pressure is constant, $p \Delta V = R \Delta T$ (this is per mole so $n=1$).
Putting all this back into the First Law
$C_v\Delta T=C_p \Delta T-R \Delta T$. Cancel $\Delta T$ and move terms around with sign changes to get
$C_p -C_v=R$.

 

[mk9898]

Thank you kuruman. Why would it be of interest to look at every mole n? Meaning, setting n =1.

 

[kuruman]

Thank you kuruman. Why would it be of interest to look at every mole n? Meaning, setting n =1.

Because if you know that the molar specific heat at constant volume is $C_v$, then the change in internal energy of 1 mole of gas when the temperature changes by $\Delta T$ will be $\Delta U = 1\times C_v \Delta T$. For $1.83$ moles of gas it will be $\Delta U = 1.83\times C_v \Delta T$ and for $39.98$ moles it will be $\Delta U = 39.98\times C_v \Delta T$. Do you see where this is going? Same goes for $C_p$.

 

[Dr Dr news]

I would venture that (cp/cv)(mixture) = 1.625

 

[mk9898]

I would venture that (cp/cv)(mixture) = 1.625

How did you get 1.625?

 

[Chestermiller]

I would venture that (cp/cv)(mixture) = 1.625

1.625 is definitely incorrect. The molar heat capacity of the mixture at constant volume is 2.33333R, and the molar heat capacity of the mixture at constant pressure is 3.33333R. So the ratio of the specific heats is 1.43.

 

[Dr Dr news]

My mistake,. Upon further review, I get cp/cv(mixture of 5 parts N2 and 1 part He) = 1.43 based on the following cv(He) = (3/2)RT, cv(N2) = (5/2)RT, cv(mix) = cv(He)/6 + cv(N2)(5/6) = (3/12 + 25/12)RT = (28/12)RT; cp(He) = (5/2)RT, cp(N2) = (7/2)RT, cp(mix) = cp(He)/6 +(5/6)cp(N2) = (5/12 + 35/12)RT = (40/12)RT; and finally cp/cv(mix) = 40/28 = 1.43 Which make sense since the specific heat ratio for air is 1.40 which is primarily diatomic N and diatomic O.

 

[Dr Dr news]

1.625 is definitely incorrect. The molar heat capacity of the mixture at constant volume is 2.33333R, and the molar heat capacity of the mixture at constant pressure is 3.33333R. So the ratio of the specific heats is 1.43.

No problem I found the error.

 

[Dr Dr news]

My mistake,. Upon further review, I get cp/cv(mixture of 5 parts N2 and 1 part He) = 1.43 based on the following cv(He) = (3/2)RT, cv(N2) = (5/2)RT, cv(mix) = cv(He)/6 + cv(N2)(5/6) = (3/12 + 25/12)RT = (28/12)RT; cp(He) = (5/2)RT, cp(N2) = (7/2)RT, cp(mix) = cp(He)/6 +(5/6)cp(N2) = (5/12 + 35/12)RT = (40/12)RT; and finally cp/cv(mix) = 40/28 = 1.43 Which make sense since the specific heat ratio for air is 1.40 which is primarily diatomic N and diatomic O.

I need to slow down. The temperatures in the above should be disregarded.