Proportion of gases in Ideal Gas Mixture? 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 $C_{C_2H_6}=0.9$



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


  Values (Curtis Phase Behaviour book p.162)
  
  
    
    $M_{CH_4}=16.04 \frac{lbm}{lbm-mol}$
    $M_{C_2H_6}=30.07 \frac{lbm}{lbm-mol}$ 
    
  
  
  What is $x_{CH_4}=\frac{n_{CH_4}}{n_{CH_4}+n_{C_2C_6}}=\frac{n_{CH_4}}{n}$?
  
  
    
    By Amagat's law and Dalton's law here,
    I know $\frac{n_j}{n}=\frac{P_j}{P}=\frac{V_j}{V}$.
    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 $m_{CH_4}=0.6m_{C_2H_6}$.
    $PV=nRT\rightarrow n=\frac{PV}{RT}$ where constant $\frac{V}{RT}$ so
    $x_{CH_4}=\frac{P_{CH_4}}{P_{CH_4}+P_{C_2C_6}}=\frac{P_{CH_4}}{P}=_{(1)}\frac{n_{CH_4}}{n}=\frac{V_{CH_4}}{V}$.
    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.
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%.