# Heat capacities of a gas mixture.

1. Jun 25, 2010

### Je m'appelle

1. The problem statement, all variables and given/known data

1 gram of Hydrogen $$H_2$$ and 1 gram of Helium $$He$$ 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 $$C_v$$ and $$C_p$$, as well as $$\gamma = \frac{C_p}{C_v}$$ of this gas mixture.

2. Relevant equations

$$n = \frac{m}{M_{molar}}$$

$$PV = nRT$$

For a monoatomic gas

$$C_v = \frac{3}{2}R$$

For a diatomic gas

$$C_v = \frac{5}{2}R$$

For both monoatomic and diatomic gas

$$C_p = C_v + R$$

$$\gamma = \frac{C_p}{C_v}$$

$$R = 8,31 \frac{J}{K.mol}$$

$$T_{kelvin} = T_{celsius} + 273$$

3. The attempt at a solution

(a)

$$P = \frac{nRT}{V}$$

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?

$$n_{He} = \frac{1}{4} = 0.25$$

$$n_{H_2} = \frac{1}{1} = 1$$

$$n_{mixture} = n_{He} + n_{H_2}$$

So,

$$P = \frac{1,25 \times 8,31 \times 300}{10} = 311,625$$

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?

$$C_v (He) + C_v (H_2) = C_v (Mixture)$$

$$C_p (He) + C_p (H_2) = C_p (Mixture)$$

$$\gamma_{mixture} = \frac{C_p (He) + C_p (H_2)}{C_v (He) + C_v (H_2)}$$

Is this correct?

Thanks in advance.

Last edited: Jun 25, 2010
2. Jun 26, 2010

### aim1732

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.

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