Physical Quantities, Units, and Conversions

Often we don’t just have numbers on their own, they are connected to physical quantities that can be measured, such as time, concentration, mass, or velocity. When describing a quantity, units are needed to define what it means physically. There is a lot of difference between 200 degrees Celsius and 200 Kelvin and 200 degrees as an angle!

All quantities share the same format:

\[ \textrm{Variable} = \textrm{Number} \times \textrm{Units} \]

For example, support we have the mass \(m\) of the product is \(m=1.2\;\textrm{g}\). Then the mass \(m\) is the variable, 1.2 is the number and g, grams, are the units.

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In Python, we can represent a physical quantity with the help of a special library for units called pint. The pint library uses the idea of a UnitRegistry to keep a track of units, and as we shall see, this can help us with unit conversions.

from pint import UnitRegistry

ureg = UnitRegistry()

m = 1.2 * ureg.g
m

1.2 gram

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It is important that we are aware of some of the rules associated with units:

• Any terms in an equation that are added or subtracted need to have the same units in order to produce an answer that makes sense. For eaxmple, the sum \(3\;\textrm{kg} + 12\;\textrm{s}\) doesn’t make physical sense.

• However, different units can be multipled together or divided to make a compound unit.

Note

If a unit has a negative power, then it is said to be “per” that unit. For example, kJ mol^-1 is pronounced as kilo-joules per mole.

Example

Speed is calculated from the equation:

\[ \textrm{speed} = \frac{\textrm{distance}}{\textrm{time}} \]

Distance is measured in metres m and time is measured in seconds s. What are the units of speed?

Solution: To do this, we replace the terms of the equation with their units.

\[ \textrm{units of speed} = \frac{\textrm{m}}{\textrm{s}} \]

This means that this speed is measured in metres per second. This can also be denoted as m/s or ms^-1.

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We can show this with pint.

distance = 1 * ureg.meter 
time = 1 * ureg.second

distance / time

1.0 meter/second

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Base Units

Note

Notice that some of the base units (kelvin and ampere) and many of the compound units (newton, joule, etc.) are named in honour of famous scientists. Even though this is the case, when we are talking about the unit, they are not written as proper nouns, i.e., there is no capital letter at the start.

There are seven fundamental quantities in all of science, which all other quantities are built from. These with their SI units are:

• Length measured in metres, m.

• Mass measured in kilograms, kg.

• Time measured in seconds, s.

• Tempearture measured in kelvin, K.

• Amount of a substance measured in moles, mol.

• Electric current measured in amperes, A.

• Luminous intensity measured in candelas cd.

The main ones we are interested in for chemistry are mass, time, temperature, amount of substance, and length. From these, we can construct a whole host of other units, such as newtons, N, and pascals, Pa.

Note

Calculating units is a good way of checking an equation has been manipulated correctly, since the units must be equal on both sides. For example, if the left hand side of the equation has units of mass per temperature, then the right has side must do as well. This is known as dimensional analysis.

Units of the Molar Gas Constant

Find the units of the molar gas constant, \(R\), using the ideal gas equation \(pV = nRT\), where \(p\) is measured in kg m^-1 s^-2, \(V\) in m³, \(n\) in mol and \(T\) in K.

Solution: To begin, we should rearrange the equation to make \(R\) the subject, as it’s easier to do this before putting the units in. So divide both sides of the equation by \(nT\) to give:

\[ \frac{pV}{nT} = \frac{nRT}{nT} \Rightarrow R = \frac{pV}{nT} \]

Now, we can substitute the units into this equation and then group the like units.

\[ \textrm{units of }R = \frac{\textrm{kg m}^{-1}\textrm{ s}^{-2}\textrm{ m}^3}{\textrm{mol K}} = \frac{\textrm{kg m}^2\textrm{ s}^{-2}}{\textrm{mol K}} = \textrm{kg m}^2\textrm{ s}^{-2}\textrm{ mol}^{-1}\textrm{ K}^{-1} \]

It might look ugly, but many physical constants have complex units.

Again, we can double check this with pint.

p = 1 * (ureg.kilogram / (ureg.meter * ureg.second ** 2))
V = 1 * (ureg.meter ** 3)
n = 1 * (ureg.mole)
T = 1 * (ureg.kelvin)

p * V / (n * T)

1.0 kilogram meter²/(kelvin mole second²)

Unit Prefixes and Scientific Notation

The numbers used in chemistry range dramatically from being very large, like Avogardo’s number, to incredibly small, such as Planck’s constant, so we use either prefixes or scientific notation to display numbers in a concise form. Take the kilogram for example, it means 1000 grams because it uses the prefix kilo, k, to mean ×10³ (which is 1000).

The SI prefixes that you are expected to know are shown in the table below.

Prefix	Symbol	Power of 10
atto	a	10-18
femto	f	10-15
pico	p	10-12
nano	n	10-9
micro	μ	10-6
milli	m	10-3
centi	c	10-2
deci	d	10-1
1	1	100
kilo	k	103
mega	M	106
giga	G	109
tera	T	1012
peta	P	1015

Example

Express the world’s population using an SI prefix to 1 significant figure.

Solution: At the time of writing the world’s population is 8 billion peopel to 1 significant figure or 8 000 000 000. This can be written as 8 × 1 000 000 000 = 8×10⁹. 10⁹ has the prefix giga, G, so the world’s population is 8 giga-people.

Scientific notation (or standard form) is another way of writing incredibly large or small numbers. Instead of writing the number to the nearest SI prefix, we write the number as a (non-zero) digit following by the decimal places all multipled by the appropriate power of ten. For example, 3112 is 3.112×10³ in standard form.

Avogardo’s Constant

Avogardo’s constant, \(N_a\), is 602 200 000 000 000 000 000 000 to 4 significant figures. Express this in standard form.

Solution: We need to first find the nearest power of 10. To do this, we count how many places are after the first digit. There are 23 places after the first digit, so the power of 10 will be 23, 10²³.

The appropraite decimal number would be 6.022, since that has a single digit followed by the decimal places.

This then makes Avogrado’s constant 6.022×10²³ to 4 s.f.

Converting Between Units

For some measurements chemists prefer to use non-SI units, because they produce nicer numbers to work with. For example, the Ångstrom (Å) is a unit of length with 1 Å = 10¹⁰ m, and hence it is a more appropriate unit when working with atomic spacings.

Important

When substituting numerical values into equations to find another quantity, we must substitute the values in their SI form. For example, when working in Ångstrom, we would first have to convert our answer to metres before substituting.

Example

The temperature outside is measured to be 95 °F. Given that Fahrenheit and Celsius are linked by the equation:

\[ \textrm{C} = \frac{5}{9} \times (\textrm{F} - 32) \]

And Celsius and kelvin are linked by the equation:

\[ \textrm{K} = \textrm{C} + 273 \]

Calculate the outside temperature in kelvin.

Solution: We need to apply 2 changes in units, first from Fahrenheit to Celsius and then from Celsius to kelvin. We know that the temperature is 95 °F. Hence, using the appropriate equation, we can find the temperature in Celsius:

\[ \textrm{C} = \frac{5}{9} \times (\textrm{F} - 32) = \frac{5}{9} \times (95-32) = \frac{5}{9} \times 63 = 35 \]

The temperature in Celsius is then 35 °C. Now all that we need to do is put this into the next equation to find the temperature in kelvin.

\[ \textrm{K} = \textrm{C} + 273 = 35 + 273 = 308 \]

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The pint library is capable of helping with unit conversions too.

Python Note

The structure of constructing with temperature quantities in pint is slightly different from other units. This is due to the fact that temperature units are non-multiplicative.

Q_ = ureg.Quantity

Q_(95, ureg.degF).to(ureg.degK)

308.15000000000003 kelvin

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Volumetic Conversions

An industrial chemist prodiuces 25×10⁵ dm³ of fertiliser in a reaction. How much is that in m³?

Solution: First, we need to find how many m³ there are per dm³. So we start with 1 m³.

\[ 1\;\textrm{m}^3 = (1\;\textrm{m})^3 \]

We know that the conversion between m and dm, from our prefixees table, is 1 dm = 0.1 m, which means that 1 m is 10 dm. Substituting this back in gives:

• Expand it out:

  \[ (1\;\textrm{m})^3 = (10\;\textrm{dm})^3 \Rightarrow 1\;\textrm{m}^3 = 10^3\;\textrm{dm}^3 \]

• Divide through by 10³:

  \[ 1\;\textrm{m}^3 = 10^3\;\textrm{dm}^3 \Rightarrow 10^{-3}\;\textrm{m}^3 = 1\;\textrm{dm}^3 \]

We have a relation that allows us to put in dm³ and get out what it is in m³.

\[ 2.5\times10^5\;\textrm{dm}^3 = 2.5\times10^5\times10^{-3}\;\textrm{m}^3 = 2.5\times10^2\;\textrm{m}^3 \]

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Finally, we achieve this with Python and pint.

(2.5E5 * (ureg.dm ** 3)).to(ureg.m ** 3)

250.00000000000006 meter³

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