Heat capacities of a gas mixture.

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 23K views
Je m'appelle
Messages
117
Reaction score
0

Homework Statement



1 gram of Hydrogen [tex]H_2[/tex] and 1 gram of Helium [tex]He[/tex] are put together into a container of 10 L in volume and at a temperature of 27°C.

(a) Find the pressure

(b) Find the molar specific heat capacities [tex]C_v[/tex] and [tex]C_p[/tex], as well as [tex]\gamma = \frac{C_p}{C_v}[/tex] of this gas mixture.

Homework Equations



[tex]n = \frac{m}{M_{molar}}[/tex]

[tex]PV = nRT[/tex]

For a monoatomic gas

[tex]C_v = \frac{3}{2}R[/tex]

For a diatomic gas

[tex]C_v = \frac{5}{2}R[/tex]

For both monoatomic and diatomic gas

[tex]C_p = C_v + R[/tex]

[tex]\gamma = \frac{C_p}{C_v}[/tex]

[tex]R = 8,31 \frac{J}{K.mol}[/tex]

[tex]T_{kelvin} = T_{celsius} + 273[/tex]

The Attempt at a Solution



(a)

[tex]P = \frac{nRT}{V}[/tex]

Now, the problem here is to find 'n' for the mixture, can I simply find the number of mols of each gas separately and then sum them up?

[tex]n_{He} = \frac{1}{4} = 0.25[/tex]

[tex]n_{H_2} = \frac{1}{1} = 1[/tex]

[tex]n_{mixture} = n_{He} + n_{H_2}[/tex]

So,

[tex]P = \frac{1,25 \times 8,31 \times 300}{10} = 311,625[/tex]

Is this correct?

(b)

In order to find out the molar heat capacity for the mixture, can I proceed just as before and work out them separately and them add them up?

[tex]C_v (He) + C_v (H_2) = C_v (Mixture)[/tex]

[tex]C_p (He) + C_p (H_2) = C_p (Mixture)[/tex]

[tex]\gamma_{mixture} = \frac{C_p (He) + C_p (H_2)}{C_v (He) + C_v (H_2)}[/tex]

Is this correct?

Thanks in advance.
 
Last edited:
Physics news on Phys.org
No the value of γ is not correct.
The best way to find it is to calculate heat capacity (as opposed to specific heat capacity) - the heat required to raise temperature of whole mixture by 1° C - for the mixture. Then divide it by the no. of moles present in the mixture.
This way you find Cp and Cv for the mixture and then γ.