# Basis of degree of freedom for monoatomic and diatomic mixture

1. May 8, 2005

### nomorevishnu

hi guys

16 grams of helium gas is mixed with 16 grams of oxygen.....
what will be the ratio Cp/Cv of the mixture.......

how to calculate it??????? i thought on the basis of degree of freedom for monoatomic and diatomic mixture......but that can take me to the answer only by approximation......how do we arrive in a a formula that could get me answer........

2. May 8, 2005

### Andrew Mason

Aren't they both diatomic?

AM

3. May 8, 2005

### nomorevishnu

how????

y do u say that these two are diatomic???

He...its completely satisfied without bonding
He is a noble gas...so y do u say it is diatomic.....

moreover someone help me solve the problem yaar........

4. May 8, 2005

### eutopia

Monatomic Gases:
Helium, Neon, Argon, Krypton, Xenon, Radon
Single atom, or monatomic, gases have the smallest Specific Heat CV.

Diatomic Gases:
Oxygen, Nitrogen, Hydrogen

5. May 8, 2005

### Andrew Mason

Of course you are right. I was seeing He and thinking H.

What is the ratio of the number of He atoms to O2 molecules?

AM

6. May 9, 2005

### nomorevishnu

hi

y isnt anyone helping me get the answer??????
please if someone can explain the problem to me...it would do a world of good to me....

please please please....the question again.....

16 grams of helium gas is mixed with 16 grams of oxygen.....
what will be the ratio Cp/Cv of the mixture.......

how to calculate it??????? i thought on the basis of degree of freedom for monoatomic and diatomic mixture......but that can take me to the answer only by approximation......how do we arrive in a a formula that could get me answer........

7. May 9, 2005

### Dr.Brain

nomorevishnu Let helium be the gas 1 and oxygen be the gas 2 .

No. Of moles of Helium: 4 = n1
" " " " oxygen: 0.5 =n2

Lamda factor for a mixture is given by:

L= n1( Cp1) + n2 (Cp2) / n1(Cv1) + n2 (Cv2)

Where Cp1/Cv1=5/3 ( for monoatomic helium)

Cp2/Cv2=7/3 for diatomic oxygen

Answer you get is: 1.62

8. May 9, 2005

### Andrew Mason

Have you worked out the proportion of numbers of atoms of He to number of molecules of O2?

AM

9. May 9, 2005

### nomorevishnu

well...i knew that equation...and the answer

but to find the effective lambda...how do we get to such an equation...any proof....its not given in Resnick and Halliday

10. May 9, 2005

### Andrew Mason

You have to go to basic principles:

$$dU = (Cp - Cv)nT = nRT$$ so:

$$C_p/C_v = \gamma = (C_v + R)/C_v = (1 + R/C_v)$$

So for the mixed gas:

$$(C_{peff} - C_{veff})n_{total}dT = n_{total}RdT$$

$$C_{peff} = (R + C_{veff})$$

(1)$$C_{peff}/C_{veff} = \gamma_{eff} = (R/C_{veff} + 1)$$

Now:

$$Vdp = (C_{vHe}n_{He} + C_{vO_2}n_{O_2})dT = C_{veff}n_{total}dT$$

(2) $$C_{veff} = (C_{vHe}n_{He} + C_{vO_2}n_{O_2})/n_{total}$$

Substitute from (2) into (1).

I get 1.64

AM

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