What are principal quantities?

A physical quantity represents the physical state of a system. It is expressed as the product of a number and unit. The basic quantities used in science are length, mass, and time. The principal quantities are the quantities that are independent of other quantities.

International system of units

A physical quantity is divided into fundamental or principal quantities and derived quantities.

According to the international system of units, there are seven principal quantities. They are given below:

Principal and derived quantities

1. Meter: The meter is the length of the path traveled by vacuum in a time interval of 1/299 second.
2. Mass: The kilogram is equal to the mass of the international prototype of the kilogram.
3. Time: The second is the duration of 9, 192, 631, 770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of the cesium-133 atom.
4. Electric current: The ampere is the constant current that would be produced between two parallel infinite conducting wires of negligible cross-section, separated by 1 m and in a vacuum which induces a force (per unit length) of $2\times {10}^{-7} N/m$.
5. Temperature: The kelvin is the fraction 1/273.16 of the triple point temperature of the water.
6. Amount of substance: The mole is the amount of substance of a system that contains as many elementary entities as 12 g of the carbon-12 isotope.
7. Luminous intensity: The candela is the luminous intensity of a source that emits monochromatic radiation of frequency 540×10¹² hertz and which has a radiance intensity of 1/683 watt per steradian.

Derived quantities are the combination of two or more primary dimensions and their units are the combination of primary units. For example, velocity can be written as length divided by time (LT^-1) and its unit is m/s. In chemical engineering practice, we encounter numerous derived quantities like force, pressure, heat, energy, enthalpy, heat capacity, mass flux, etc. The understanding of these quantities and representing them in the correct dimension and unit is very essential in designing various processes. The conversion of units to different systems of units like metric system or MKS system, SI system, English or FPS system is also important.

Equation of state (EOS)

In chemical engineering, the thermodynamic state of a single-phase, homogeneous substance is specified by thermodynamic properties like temperature (T), pressure (P), and volume (V). There exists a unique relation between P, V and T. Analytically it can be expressed as 

f (P, V, T) = 0

This relation is called the PVT equation of state. The simplest EOS is for ideal gases which are given by the ideal gas law PV=RT.

The equation of state can be solved to get one of the unknown quantities from P, V,  and T when the other two values are given.

When V is considered as a function of P and T, V=V (T, P) and it gives a relation,

$dV={\left(\frac{\partial V}{\partial T}\right)}_P\partial T +{\left(\frac{\partial V}{\partial P}\right)}_T\partial P$

The partial derivative in this equation is related to two properties - volume expansivity and isothermal compressibility.

1. Volume expansivity,  $\beta =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P$
2. Isothermal compressibility, $K=\frac{-1}{V}{\left(\frac{\partial V}{\partial P}\right)}_T$

For incompressible fluids, Ƙ is zero. For a small change in temperature and pressure, Ƙ is constant. The integration of the above equation gives,

$In\frac{V_2}{V_1}=\beta (T_2-T_1)-K(P_2-P_1)$

Equation of state for ideal gases

An ideal gas state is a hypothetical state of matter. It is composed of real molecules which have negligible molecular volume and no intermolecular forces.

For 1mol the ideal gas equation of state is PV = RT where R is the ideal gas constant which is equal to 8.314 J/mol K. This equation is used to calculate property change for real gases. Gases like hydrogen, helium, nitrogen, and oxygen behave like an ideal gas at room temperature.

Equation of state for real gases

Real gases to become ideal, the molecular interaction should be negligible. They follow ideal behavior at zero pressure. Many equations of state have been proposed to study the behavior of real gases.

Virial equation

Experimentally it has been observed that the ratio $\frac{PV}{RT}$ can be expressed as a power series of P or $\frac{1}{V}$. The ratio of the volume of real gas (V) to the volume of a gas if behaving ideally ($\frac{RT}{P}$) is known as the compressibility factor, $Z=\frac{PV}{RT}$.

Virial equation is given by

$Z=1+Bʹ P+Cʹ P2+Dʹ P3 +\dots \dots$

Or

$Z= 1+\frac{B}{V}+ \frac{B}{V^2}+\dots$

The constants B, C, D, Bʹ, Cʹ, Dʹ are called Virial constants. At low or moderate pressure, these equations can be truncated after the second term to calculate various unknown quantities.

Cubic equation of state

An EOS to represent the PVT behavior of liquids and vapors must cover a wide range of temperatures, pressures, and molar volumes. Polynomial equations that are cubic in the molar volume are simpler and suitable for many purposes when studying PVT behavior. Cubic equations are capable of representing both liquid and vapor behavior.

1. Van der Waals equation

To explain the P-V-T behavior of real gases van der Waals proposed the following cubic equation

$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT$

Here, a and b are van der Waals constants. The term $\frac{a}{V^2}$ accounts for the attractive forces between molecules. The constant b accounts for the finite size of molecules, which makes the volume larger than in the ideal-gas state. If values of a and b for a particular fluid are given, we can calculate P as a function of V for various values of T. The constants in the van der Waals equation can be calculated using limiting conditions which state that at critical points The P-V curve at constant temperature shows the point of inflection.

 ${\left(\frac{\partial P}{\partial V}\right)}_{T_c}=0 and {\left(\frac{{\partial }^{2}P}{\partial V^2}\right)}_{T_c}=0$

The constants are given by $a=\frac{27 R^2 {T^2}_C}{64P_C} and b=\frac{R{T}_C}{8P_C}$.

Here $T_C$ and $P_C$ are the critical temperature and pressure which are the highest temperature and pressure at which pure material can exist in vapor-liquid equilibrium.

2. Redlich-Kwong (RK) equation

The Redlich-Kwong (RK) equation is a two-parameter cubic equation of state that is widely used in engineering applications.

$P=\frac{RT}{V-b}-\frac{a}{T^{0.5}V\left(V-b\right)}$

The constants are given by

$a=\frac{0.427 R^2 {T^{2.5}}_C}{P_C}$ and $b=\frac{0.0867R{T}_C}{8P_C}$

Context and Applications

Principal quantities are the base quantities of measurement for various physical quantities.  Understanding principal and derived quantities, their units, and conversion to other unit systems are very important in chemical process calculations, equipment design, and thermodynamics. Equation of state plays an important role in describing the fluid behavior in thermodynamics, chemical process calculations, design, etc. The equation of state helps to evaluate various properties of pure substance and mixture under given conditions. This topic can be applied in the below-mentioned courses,

• Bachelor in Engineering (Chemical, Biotechnology, Mechanical)
• Master in Engineering ( Chemical, Biochemical)
• Bachelor in Science (Physics, Chemistry)

Related Concepts

• P-V-T behavior
• Dimensional analysis
• Thermodynamics
• Thermal equilibrium

Practice Problems

1. Calculate the molar volume of air at 300 K and 1 bar assuming air behaves as an ideal gas.

1. $49\times {10}^{-2} m^3/mol$
2. $2.49\times {10}^{-2} l/mol$
3. $5\times {10}^{-2} m^3/mol$
4. $2.49\times {10}^{-2} c{m}^3/mol$

Correct answer: Option a 

Solution:  From Ideal gas law V=RT/P $=8.314\times 300/{10}^5$ = $2.49\times {10}^{-2} m^3/mol$

Note: $1 bar={10}^5 N/m^2$

2. Find the pressure exerted by 1kmol of $C{O}_2$ if it occupies $0.4 m^3$ at 313 K using the van der Waals equation. Given $a=0.365 Nm/mo{l}^2$ and $b=4.28\times {10}^{-5} m^3/mol$

1. $50 \times {10}^5 N/m^2$
2. $50 \times {10}^5 bar$
3. $100 \times {10}^5 N/m^2$
4. $100 \times {10}^5 bar$

Correct answer: Option a 

Solution: Substitute the given values in the van der Waals equation
$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT$

Where $V=0.4\times {10}^{-3} m^3/mol$ and calculate $P=50\times {10}^5 N/m^2$

3. The density of 1mol of an ideal gas can be calculated using which of the following equations?

1. $Density=\frac{RT}{V}$
2. $Density= \frac{PM}{RT}$
3. $Density=\frac{P}{RT}$
4. $Density= \frac{Pn}{RT}$

Correct answer: Option b

Solution: For ideal gas PV=RT means, volume $=\frac{P}{RT}$

Mass of 1 mol. of gas= Molecular weight=M

Density of 1mol of ideal gas$=\frac{Mass}{Volume}=\frac{PM}{RT}$

4. What does the term $\frac{a}{V^2}$ in the van der Waals equation represent?

1. It accounts for the attractive forces between molecules.
2. It accounts for the finite size of molecules.
3. It accounts for the kinetic energy of molecules.
4. It accounts for the repulsive force between molecules.

Correct answer: Option a

Solution: The term $\frac{a}{V^2}$ in the van der Waals equation accounts for the attractive forces between molecules.

5. Which of the following is the Virial equation?

1. $Z=1+Bʹ P2+Cʹ P3+Dʹ P4 +\dots \dots$
2. $Z=1+\frac{B}{V}+\frac{B}{V^2}+\frac{B}{V^2}+\cdots$
3. $(P+\frac{a}{V^2})(V-b)=RT$
4. Z =1+Bʹ V+Cʹ V²+Dʹ V³+……

Correct answer: Option b

Solution: The compressibility factor $Z=\frac{PV}{RT}$ can be expressed as a power series of P or $\frac{1}{V}$. Hence, the correct option is $Z=1+\frac{B}{V}+\frac{B}{V^2}+\frac{B}{V^2}+\cdots$