A certain gas obeys the van der Waals equation with a = 0.50 m6 Pa mol−2. Its molar volume is found to be 5.00 × 10–4 m3 mol−1 at 273 K and 3.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure?

Respuesta :

Answer:

a

The value is [tex]b  =  4.61 *10^{-5} \  m^3 /mol[/tex]

b

[tex]Z =  0.66[/tex]

Explanation:

Generally the pressure of the gas  is mathematically represented as

     [tex]P  =  \frac{RT}{V_m -b}  - \frac{a}{(V_m)^2}[/tex]

Substituting [tex]8.314 \  m^3\cdot Pa\cdot K^{-1}\cdot mol^{-1}[/tex] for R , 273K for  T  , [tex]5.00 * 10^{-4} m^3[/tex] for  [tex]V_m[/tex] ,  [tex]0.50 \ m^6 \cdot Pa\cdot  mol^{-2}[/tex] for a  and  [tex]3.0 MPa =  3.0*10^{9} \ Pa[/tex] for P

We have

     [tex]3*10^6 =  \frac{8.314 * 273}{5.00*10^{-4} - b}  - \frac{0.50}{(5.00*10^{-4})^2 }[/tex]

=>  [tex]b  =  4.61 *10^{-5} \  m^3 /mol[/tex]

     

Generally the compression factor is mathematically represented as

       [tex]Z =  \frac{P* V_m }{RT}[/tex]

=>   [tex]Z =  \frac{3.0 *10^{6}* 5.0 *10^{-4} }{8.314 *273}[/tex]

=>     [tex]Z =  0.66[/tex]