Finding the compressibility factor (Z)

In summary, when thermodynamic pressures and temperatures are high enough, hydrogen becomes a gas instead of an element, leading to a different value for Z. This value is found by dividing the pressure of hydrogen (P) by its volume at standard temperature and pressure (V), then multiplying by the ideal gas constant (R).
  • #1
awyea
3
0

Homework Statement


Determine the volume, in m^3, occupied by 20 kg of hydrogen (H2) at 1170 kPa, 2220°C.

Homework Equations



Z=pv/rt, Pr=P/Pc, Tr=T/Tc, and for hydrogen M = 2.016 (kg/kmol) Tc = 33.2 (K) Pc = 13.0 bar Zc=pc*vc/(RTc)


The Attempt at a Solution


I know if I find Z then the problem is a done deal. I'm just confused about how to use Pr,Tr, and the table to find Z. I found Pr=.9 and Tr=1.59 but is there a formula for Z?

Thanks in advance for the help (I'm not a fan of Thermo so far and my Profs at my old college spoiled me with great teaching)
 
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  • #2
Isn't it just an ideal gas question?
 
  • #3
Sadly no, Z is the factor that relates PV/RT of Hydrogen to the PV/RT of and ideal gas. Hydrogen in this problem, because of the high temperatures and pressure, is not an ideal gas.
 
  • #4
awyea said:

Homework Statement


Determine the volume, in m^3, occupied by 20 kg of hydrogen (H2) at 1170 kPa, 2220°C.

Homework Equations



Z=pv/rt, Pr=P/Pc, Tr=T/Tc, and for hydrogen M = 2.016 (kg/kmol) Tc = 33.2 (K) Pc = 13.0 bar Zc=pc*vc/(RTc)


The Attempt at a Solution


I know if I find Z then the problem is a done deal. I'm just confused about how to use Pr,Tr, and the table to find Z. I found Pr=.9 and Tr=1.59 but is there a formula for Z?

Thanks in advance for the help (I'm not a fan of Thermo so far and my Profs at my old college spoiled me with great teaching)

Something is sadly amiss. You have said that you found Tr = 1.59, after also saying that T = 2220+273 K and Tc = 33.2K. But when I calculate using the equation that you have supplied

Tr = T/Tc

I get Tr = 2493/33.2 , which is roughly 78, not 1.59! Your value for pr seems about right, although you will need to carry at least 1 more significant figure, and preferably 2 more to get a result with any precision.

The other point is that I am not seeing an equation for whichever approach you are using to approximate the ideal gas equation, if that is not good enough. Van der Waals equation? Virial theorem calculation? or what?
 
  • #5
Ok, I feel really bad about this, the T value is -220 °C, which makes the Tr come out to 53.0K/33.2k=1.59. Pr=.900 with the correct number of sig. figs. After reading the question twice more, looking for something about how to approximate Z, i finally found the attached chart hiding in the back of the book. This makes it look like Z should be .95, although it seems like a very inaccurate way to find it. If this is true, v=(8.314m3Pa/(k*mol)*53K*.95/(11.7x10^3 Pa)=.0358m3/mol
Using dimensional analysis, .0358m3/mol / .00202 kg/mol * 20kg = 354 m3, and it looks like I could have done this problem without wasting everyone's time. Sorry about that.
 

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1. What is the compressibility factor (Z)?

The compressibility factor (Z) is a dimensionless ratio that represents the deviation of a real gas from an ideal gas. It takes into account the effects of intermolecular forces and non-ideal behavior of gases.

2. How is the compressibility factor (Z) calculated?

The compressibility factor (Z) can be calculated using the following equation: Z = PV/RT, where P is the gas pressure, V is the gas volume, R is the gas constant, and T is the temperature.

3. What is the significance of the compressibility factor (Z) in gas behavior?

The compressibility factor (Z) is important in understanding the behavior of real gases, as it helps to predict their properties such as density, pressure, and volume. It also plays a role in determining the phase behavior of gases.

4. How does the compressibility factor (Z) vary with temperature and pressure?

The compressibility factor (Z) varies with temperature and pressure, as it is affected by the intermolecular forces between gas molecules. At low temperatures and high pressures, Z is typically greater than 1, indicating non-ideal behavior. At high temperatures and low pressures, Z is usually less than 1, showing that the gas behaves more ideally.

5. What are some practical applications of the compressibility factor (Z)?

The compressibility factor (Z) has various practical applications in the fields of chemistry, engineering, and materials science. It is used in the design and operation of gas pipelines, as well as in the development of new materials for gas storage. It also plays a role in the production and purification of natural gas and other industrial gases.

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