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Suppose methane ($CH_4$) and ethane ($C_2H_6$) in a container. The specific gravity is $r_g=0.6$. I think this means $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$. Now I am trying to understand the question.

Question: what is the proportion of gases?

Solution according to my teacher: $C_{CH_4}=0.1$ and $C_{C_2H_6}=0.9$

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My Solution: $C_{CH_4}=24.2\%$ and $C_{C_2H_6}=75.8\%$

Values (Curtis Phase Behaviour book p.162)

  1. $M_{CH_4}=16.04 \frac{lbm}{lbm-mol}$

  2. $M_{C_2H_6}=30.07 \frac{lbm}{lbm-mol}$

What is $x_{CH_4}=\frac{n_{CH_4}}{n_{CH_4}+n_{C_2C_6}}=\frac{n_{CH_4}}{n}$?

  1. By Amagat's law and Dalton's law here, I know $\frac{n_j}{n}=\frac{P_j}{P}=\frac{V_j}{V}$.

  2. Specific gravity $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$ where $\rho=\frac{m}{V}$ and constant $V$ so $0.6=\frac{m_{CH_4}}{m_{C_2H_6}}$ so $m_{CH_4}=0.6m_{C_2H_6}$.

  3. $PV=nRT\rightarrow n=\frac{PV}{RT}$ where constant $\frac{V}{RT}$ so $x_{CH_4}=\frac{P_{CH_4}}{P_{CH_4}+P_{C_2C_6}}=\frac{P_{CH_4}}{P}=_{(1)}\frac{n_{CH_4}}{n}=\frac{V_{CH_4}}{V}$.

  4. By $M_i=\frac{n_i}{m_i}$ so $n_i=M_i m_i$ where $n_{CH_4}=M_{CH_4}m_{CH_4}$

so

$\frac{n_{CH_4}}{n}=\frac{M_{CH_4}m_{CH_4}}{M_{CH_4}m_{CH_4}+M_{C_2H_6}m_{C_2H_6}}=_{(2)}\frac{M_{CH_4}0.6m_{C_2H_6}}{M_{CH_4}0.6m_{C_2H_6}+M_{C_2H_6}m_{C_2H_6}}=\frac{0.6M_{CH_4}}{M_{CH_4}0.6+M_{C_2H_6}}=0.242455$

$\rightarrow\frac{n_{C_2H_6}}{n}=1-\frac{n_{CH_4}}{n}=0.757545$

where the results are not 0.1 and 0.9.

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Your interpretation of specific gravity is incorrect:

Specific gravity (SG) refers to the ratio between a given density and that of a reference. Usually it is defined for liquids with respect to water and gasses with respect to air so in this case

$0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

where $n_i$ are proportions by mole factor. This is the incorrect assumption

$0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$.

Gasses are always a little dicier since the reference is much more important with changes in temperature and pressure affecting the resulting density.

Example

The density form of the ideal gas laws goes like:

$\rho = \text{mass / volume = molecular weight * pressure / constant / temperature}$.

Therefore for constant temperature and pressure, the specific gravity is equal to the ratio of molecular weights. This follows because $\frac{\rho_1}{\rho_2} = \frac{MW_1}{MW_2}$ at constant conditions.

Equations and Solution

$M_{\text{air}} = 28.97$ (I got $28.84$ using the simple $0.21*32+0.79*28$ and this agrees with this)

$\text{SG}=0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

$1=n_{CH_4}+n_{C_2H_6}$

so by the first equation and the second equation the density of the mixture is $0.6 * 28.97 = 17.38$. The third equation holds only if $0.9*16 + 0.1*30 = 17.4$ so

methane is 90% and ethane is 10%.

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    $\begingroup$ Thank you! I made it easier to read with LaTex/structuring keeping the original content the same. I hope people will accept my edits so easier to read the key points in the future +1. Welcome to SE! $\endgroup$ – hhh Nov 30 '14 at 11:21
  • $\begingroup$ I would remove the part Example and leave only the Equations and Solution part -- even though I read it many times I cannot understand how it is related to this question. Or perhaps you can clarify it? $\endgroup$ – hhh Nov 30 '14 at 11:38
  • $\begingroup$ I was trying to emphasize how to get from the specific gravity to the ratio of molecular weights via the ideal gas law. $\endgroup$ – Byron Wall Dec 1 '14 at 15:02

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