# Volume using Compressibility Factor

• MechE2015
In summary, the problem is to find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z. The relevant equations are PV = ZnRT and PV = mRT. The given values for Tr, Pr, and R are 0.9561, 0.496, and 0.2964 kJ/kg K respectively. Using the equation V = ZnRT/P, with the appropriate units, the volume is found to be too low. To solve this, make sure to use the correct units for R and consider using the equation PV = mRT if using kj/kg K.
MechE2015

## Homework Statement

Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z

## Homework Equations

Method to Solve for Z, Using Tr and Pr

PV = ZnRT

## The Attempt at a Solution

Tr = 0.9561
Pr = 0.496
Z found to be approx. 0.75
R given on a table at 0.2964 kJ/kg K

From here, I should have all information needed, I was just confused on units at this point.
V = ZnRT/P
V = (0.75)*(n)*(0.2964 kJ/kg K)*(270 K) / (2500 kPa)

At this point, my main questions are (probably very simple):

What is n, and why? My best guess: (2000 g)/(28 g/mol) = 71.43 mol

What are units for P? My best guess: atm, -> 2500 kPa*24.67 atm/kPa = 61675 atm

Using these numbers, I get Volume to be way too low, corrections needed somewhere. Help appreciated.

MechE2015 said:

## Homework Statement

Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z

## Homework Equations

Method to Solve for Z, Using Tr and Pr

PV = ZnRT

## The Attempt at a Solution

Tr = 0.9561
Pr = 0.496
Z found to be approx. 0.75
R given on a table at 0.2964 kJ/kg K

From here, I should have all information needed, I was just confused on units at this point.
V = ZnRT/P
V = (0.75)*(n)*(0.2964 kJ/kg K)*(270 K) / (2500 kPa)

At this point, my main questions are (probably very simple):

What is n, and why? My best guess: (2000 g)/(28 g/mol) = 71.43 mol

What are units for P? My best guess: atm, -> 2500 kPa*24.67 atm/kPa = 61675 atm

Using these numbers, I get Volume to be way too low, corrections needed somewhere. Help appreciated.

Take a look at your units. If you're using kj/kg K for R, you should use equation:
PV = mRT, where R is the gas constant for that gas and m is the mass in kilograms. If you use PV = nRuT (Where Ru is the universal gas constant), then you'll need to use the number of moles.

Also, use the fact that 1 Pa*m^3 = 1 J.

## 1. What is compressibility factor?

Compressibility factor is a dimensionless quantity that relates the actual volume of a gas to the volume it would occupy at standard conditions of temperature and pressure. It is represented by the symbol Z and is a measure of how easily a gas can be compressed or expanded.

## 2. How is compressibility factor calculated?

The compressibility factor is calculated by dividing the actual volume of a gas by the ideal gas volume at the same conditions of temperature and pressure. This can be expressed as Z = Vactual/Videal.

## 3. What is the significance of compressibility factor in gas behavior?

Compressibility factor is an important factor in understanding the behavior of gases. It is used to determine the deviation of a gas from ideal gas behavior and can provide insights into the intermolecular forces and interactions within a gas. It also affects the compressibility and density of a gas, which can impact its flow and behavior in different conditions.

## 4. How does compressibility factor change with temperature and pressure?

The compressibility factor of a gas varies with temperature and pressure. At low temperatures and high pressures, gases tend to have higher values of Z, indicating stronger intermolecular forces. At high temperatures and low pressures, gases behave more like ideal gases and have lower values of Z. At the critical point, the compressibility factor is equal to 1.

## 5. How can compressibility factor be used in engineering and industry?

Compressibility factor is a crucial factor in the design and operation of gas pipelines, storage tanks, and other engineering systems that involve the transportation and handling of gases. It is also used in the development of equations of state that can accurately predict gas behavior under different conditions. Additionally, compressibility factor is important in the production and refinement of natural gas and other gases in the industry.

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