Answer:
The ideal molar volume is [tex]\frac{V}{n} =V_z= 0.001095 \ m^3/mol[/tex]
The Z factor is [tex]Z = 0.09997[/tex]
The real molar volume is [tex]\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}[/tex]
Explanation:
From the question we are told that
The pressure is [tex]P = 23 \ bar = 23 *10^5 Pa[/tex]
The temperature is [tex]T = 30 ^ oC = 303 \ K[/tex]
According to the ideal gas equation we have that
[tex]PV = nRT[/tex]
=> [tex]\frac{V}{n}=V_z= \frac{RT}{P}[/tex]
Where [tex]\frac{V}{n }[/tex] is the molar volume and R is the gas constant with value
[tex]R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}[/tex]
substituting values
[tex]\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}[/tex]
[tex]\frac{V}{n} =V_z= 0.001095 \ m^3/mol[/tex]
The compressibility factor of the gas is mathematically represented as
[tex]Z = \frac{P * V_z}{RT}[/tex]
substituting values
[tex]Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}[/tex]
[tex]Z = 0.09997[/tex]
Now the real molar volume is evaluated as
[tex]\frac{V_r}{n} = V_k= \frac{Z * RT }{P}[/tex]
substituting values
[tex]\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}[/tex]
[tex]\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}[/tex]
