Chapter 5.2 - Charles' Law

In the previous section, we saw the basic details about Boyle's law. In this section, we will see Charles' law. We will also learn about the Kelvin temperature scale

• Charles and Gay Lussac worked independently, trying to improve the hot air balloon technology
• They discovered a basic fact. It can be written as:
■ If the pressure and mass of a gas remains constant, then:
    ♦ The volume of that gas will increase if it’s temperature is increased
    ♦ The volume of that gas will decrease if it’s temperature is decreased

■ Their findings were published as: Charles’ Law
■ The law states that:
At constant pressure, the volume of a fixed mass of a gas is directly proportional to it’s absolute temperature
• We can write an explanation in 30 steps
(While writing the explanation, we will also learn what ‘absolute temperature’ is)

1. Take a sample of a gas. It must be at a temperature of 0 ^oC  

• Note down the number of moles present in it

Some examples:

• If the gas is N₂, and if ‘m’ grams of N₂ is present in that sample, then:

    ♦ There will be $\mathbf\small{\rm{\frac{m}{28}}}$ moles of N₂ in that sample

• If the gas is CO₂, and if ‘m’ grams of CO₂ is present in that sample, then:

    ♦ There will be $\mathbf\small{\rm{\frac{m}{44}}}$ moles of CO₂ in that sample

2. Note down the pressure p of the sample

3. Measure the initial volume V₀ of the sample

4. Measure the initial temperature t₀ of the sample and confirm that it is 0 ^oC

    ♦ For our present case, temperatures must be measured in the Celsius scale. Not in the Kelvin scale

5. Increase the temperature of the sample by 1 ^oC

    ♦ So the final temperature (t₁) of the sample will be given by: t₁ = (t₀+1) = (0+1) = 1 ^oC

• During this rise in temperature, the pressure must remain at the value noted down in (2) 

• Measure the final volume V₁

• Charles observed that: $\mathbf\small{\rm{(V_1-V_0)=\frac{1}{273.15}\times V_0}}$

• This observation can be explained in 3 steps:

(i) (V₁-V₀) obviously, is the increase in volume when the temperature was increased by 1 ^oC

(ii) V₀ is the volume of the gas when it’s temperature is 0 ^oC

(iii) So we can write:

When the temperature increases by 1 ^oC, the volume increases by [$\mathbf\small{\rm{\frac{1}{273.15}}}$ times of V₀]

6. Increase the temperature of the sample by another 1 ^oC

    ♦ So the final temperature (t₂) of the sample will be given by: t₂ = (t₀+2) = (0+2) = 2 ^oC

• During this rise in temperature, the pressure must remain at the value noted down in (2) 

• Measure the final volume V₂

• Charles observed that: $\mathbf\small{\rm{(V_2-V_0)=\frac{2}{273.15}\times V_0}}$

• This observation can be explained in 3 steps:

(i) (V₂-V₀) obviously, is the increase in volume when the temperature was increased by 2 ^oC

(ii) V₀ is the volume of the gas when it’s temperature is 0 ^oC

(iii) So we can write:

When the temperature increases by 2 ^oC, the volume increases by [$\mathbf\small{\rm{\frac{2}{273.15}}}$ times of V₀]

7. Increase the temperature of the sample by another 1 ^oC

    ♦ So the final temperature (t₃) of the sample will be given by: t₃ = (t₀+3) = (0+3) = 3 ^oC

• During this rise in temperature, the pressure must remain at the value noted down in (2) 

• Measure the final volume V₃

• Charles observed that: $\mathbf\small{\rm{(V_3-V_0)=\frac{3}{273.15}\times V_0}}$

• This observation can be explained in 3 steps:

(i) (V₃-V₀) obviously, is the increase in volume when the temperature was increased by 3 ^oC

(ii) V₀ is the volume of the gas when it’s temperature is 0 ^oC

(iii) So we can write:

When the temperature increases by 3 ^oC, the volume increases by [$\mathbf\small{\rm{\frac{3}{273.15}}}$ times of V₀]

8. So it is clear that:

For each ‘temperature increase’ of 1 ^oC, a ‘volume increase’ of [$\mathbf\small{\rm{\frac{1}{273.15}}}$ times of V₀] will take place

■ So we can write the general form:

If the ‘temperature increase’ is t ^oC, the ‘volume increase’ will be [$\mathbf\small{\rm{\frac{t}{273.15}}}$ times of V₀]

9. If after the ‘temperature increase’ of t ^oC, the final volume is V_t, we can write:

The increase in volume = (V_t-V₀)

• But from (8), this 'volume increase' is [$\mathbf\small{\rm{\frac{t}{273.15}}}$ times of V₀]

• So we can equate the two: $\mathbf\small{\rm{V_t-V_0=\frac{t}{273.15} \times V_0}}$

$\mathbf\small{\rm{\Rightarrow Vt=\left(\frac{t}{273.15} \times V_0 \right)+V_0}}$

• V₀ and 273.15 are constants. We will bring them together. So we get:

$\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)t+V_0}}$

10. In the above equation:

    ♦ V₀ and 273.15 are constants

    ♦ V_t and t are variables

• So we plot a V_t vs t graph

    ♦ We can plot volumes along the y-axis

    ♦ We can plot temperatures along the x-axis

• An example is shown in fig.5.10 below:

Fig.5.10

■ In this graph, all the readings were taken when the pressure of the sample was at a constant value. If there is any change in that pressure, while any of the readings are taken, we will not get this shape

■ Also, if there is any change in the 'number of moles in the sample', we will not get this shape
11. The shape of the graph in fig.5.10 is: straight line

• This is expected because:

    ♦ [$\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)t+V_0}}$] is the equation of a straight line

    ♦ It is of the form: [y = mx+c] that we commonly see in analytical geometry classes

          ✰ m is the slope of the line. It is a constant

          ✰ c is also a constant

12. In the fig.5.10, we see that, the line slopes downwards towards the left

    ♦ That means, when we move towards the left, the volume decreases

    ♦ That means, when temperature decreases, volume also decreases

■ So we will be interested in finding this:

The volume V_t when the temperature decreases to zero ^oC

    ♦ We can find it using both graphical and analytical methods

■ First we will see the graphical method. It can be written in 3 steps:  

(i) Extend the line towards the left so that it cuts the y-axis

    ♦ This is shown in fig.5.11 below

    ♦ The extension towards the left is indicated by the red dashed line

(ii) Mark 'A', the point of intersection with the y-axis

(iii) y-coordinate of A is the 'volume when temperature is zero'

• This is because, for all points on the y-axis:

    ♦ The x-coordinate will be zero

          ✰ 'Zero x-coordinate' indicates zero temperature

    ♦ The y-coordinate will give the volume

          ✰ This is because, y-axis is the volume axis

Fig.5.11

■ Next we will see the analytical method. It can be written in 3 steps:

(i) We have the equation of the line: $\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)t+V_0}}$

(ii) We want to find the volume at ‘zero temperature’. So we put t = 0

• We get: $\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)\times 0+V_0}}$

$\mathbf\small{\rm{\Rightarrow V_t=V_0}}$

• Indeed, V₀ is the volume when the temperature is zero ^oC

(iii) Let us compare [$\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)t+V_0}}$] and [y = mx+c]

• We see that:

    ♦ V₀ corresponds to c

    ♦ From analytical geometry classes, we know that: c is the y-intercept

          ✰ Indeed in our present case, V₀ is the y-intercept

13. We have seen that: When temperature decreases, volume decreases

■ So we will be interested in finding this:

The temperature at which the volume decreases to zero 

    ♦ We can find it using both graphical and analytical methods

■ First we will see the graphical method. It can be written in 5 steps:  

(i) Extend the line towards the left far beyond A

    ♦ Extend it till it meets the x-axis

    ♦ This is shown in fig.5.11 above

    ♦ The extension towards the left is indicated by the red dashed line

(ii) Mark 'B', the point of intersection with the x-axis

(iii) x-coordinate of B is the 'temperature when volume is zero'

• This is because, for all points on the x-axis:

    ♦ The y-coordinate will be zero

          ✰ 'Zero y-coordinate' indicates zero volume

    ♦ The x-coordinate will give the temperature

          ✰ This is because, x-axis is the temperature axis

■ Next we will see the analytical method. It can be written in 3 steps:

(i) We have the equation of the line: $\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)t+V_0}}$

(ii) We want to find the temperature at ‘zero volume’. So we put V_t = 0

• We get: $\mathbf\small{\rm{0=\left(\frac{V_0}{273.15} \right)t+V_0}}$

$\mathbf\small{\rm{\Rightarrow V_0=\left(\frac{-V_0}{273.15} \right)t}}$

$\mathbf\small{\rm{\Rightarrow 1=\left(\frac{-1}{273.15} \right)t}}$

$\mathbf\small{\rm{\Rightarrow t=-273.15}}$

(iii) ‘-273.15’ is a point on the x-axis

    ♦ On the x-axis, the temperatures are marked in ^oC

    ♦ So ‘-273.15’ is a temperature in ^oC

■ Thus we can write:

When the temperature of a sample of gas becomes -273.15 ^oC, the volume of that sample becomes zero

    ♦ In practice, most gases become liquid before -273.15 ^oC is reached

14. The temperature (-273.15 ^oC) at which all gases would have zero volume is a well defined point

    ♦ The molecules in all substances will cease to move at that temperature

    ♦ That means, at that temperature, kinetic energy will be zero

■ A lower temperature than -273.15 c is impossible to attain

15. Such a well defined point can be used as a bench mark for a new scale

■ This new scale is called Kelvin temperature scale

In this scale, the starting point is the 'lowest possible temperature'

16. The lowest possible temperature is: -273.15 ^oC

• But we get the ‘number -273.15’ only when we use the Celsius scale

• We have to give a 'suitable number' for this ‘lowest possible temperature’ in the kelvin scale

■ ‘0’ will be a 'suitable number' because, it is the starting point

17. So we can write an interesting comparison between the two scales

• It can be written in 2 steps:

(i) In Celsius scale, -273.15 ^oC can be considered as the starting point

    ♦ All readings will be higher than -273.15 ^oC

• All readings will be higher than -273.15 ^oC, simply because, it is impossible to attain temperatures lower than -273.15 ^oC

(ii) In Kelvin scale, 0 K is considered as the starting point

    ♦ All readings will be higher than 0 K

• All readings will be higher than 0 K, simply because, it is impossible to attain temperatures lower than 0 K

18. From the above comparison, it is clear that:

-273.15 ^oC in the Celsius scale corresponds to 0 K in the kelvin scale

• Let us consider an interesting scenario. It can be written in 4 steps:

(i) An object is at the ‘lowest possible temperature’

(ii) There are two thermometers to measure it’s temperature

    ♦ Thermometer A  and Thermometer B

          ✰ A measures temperature in the Celsius scale

          ✰ B measures temperature in the Kelvin scale

(iii) When we use the two thermometers on the object, we will get the following readings:

    ♦ The reading in A will be -273.15 c

    ♦ The reading in B will be 0 K

(iv) In the Kelvin thermometer, there will be no '-ve markings' because, we will not find any temperatures lower than 0 K

19. So it can be confirmed:

• This correspondence is indicated by points B and C in the fig.5.12 below

• The fig.5.12 shows a few more points: A, D, E and F

• Learning the ,significance of all the points' in the fig., will help us to get a good understanding about the Kelvin scale

• Detailed steps are given in a separate section 5.2(a)

Fig.5.12

20. After visiting section 5.2(a), we have a basic understanding about the Kelvin scale. Also, we have the following information:

■ To convert ^oC into K:

    ♦ K = 273.15 + C

    ♦ Some examples:

          ✰ 120 ^oC = (273.15 + 120) = 393.15 K

          ✰ -135 ^oC = (273.15 - 135) = 138.15 K 

■To convert K into ^oC:

    ♦ C = K - 273.15 

    ♦ Some examples:

          ✰ 320 K = (320 - 273.15) = 46.85 ^oC

          ✰ 142 K = (142 - 273.15) = -131.15 ^oC.

21. Now we go back to our main discussion

• In step (9), we derived the equation: $\mathbf\small{\rm{V_t=\left(\frac{V_0}{273.15} \right)t+V_0}}$

• This can be rearranged as: $\mathbf\small{\rm{V_t=V_0 \left[\left(\frac{t}{273.15} \right)+1 \right]}}$

$\mathbf\small{\rm{\Rightarrow \frac{V_t}{V_0}= \left[\frac{t+273.15}{273.15} \right]}}$

22. Consider the numerator on the right side: (t+273.15)

• In step (20) above, we saw that:

If 273.15 is added to a temperature in ^oC, that temperature will get converted into K temperature

• So the numerator is the 'K equivalent' of t ^oC

• We will call it: T_t K

23. Consider the denominator in the right side: 273.15

• It is 'K equivalent' of 0 ^oC

• We will call it: T₀ K

24. The result in (21) will become: $\mathbf\small{\rm{\frac{V_t}{V_0}= \frac{T_t}{T_0}}}$

25. Recall that in step (4), we specifically said that, all temperatures should be measured in Celsius scale

■ But now, all those temperatures are automatically converted into the Kelvin scale

26. The result in (24) can be rearranged as:

$\mathbf\small{\rm{\frac{V_t}{T_t}= \frac{V_0}{T_0}}}$

■ That means:

    ♦ The ratio of 'volume at a particular temperature' to 'that temperature'

    ♦ is equal to

    ♦ The ratio of 'volume at absolute zero temperature' to 'the absolute zero temperature'

27. Both V₀ and T₀ are constants

• So we can put: $\mathbf\small{\rm{\frac{V_t}{T_t}= k_2}}$

• Thus we get: $\mathbf\small{\rm{V_t= k_2 \times T_t}}$

• That means, volume is proportional to temperature

28. Based on this proportionality, we can write:

• V₁ = k₂ × T₁

• V₂ = k₂ × T₂

• So on . . . 

■ That is: $\mathbf\small{\rm{\frac{V_1}{T_1}= \frac{V_2}{T_2}}}$

    ♦ This is the mathematical form of Charles law

    ♦ For this equation to be valid, pressure and mass must remain a constant

29. In step (27), we have a relation between volume and temperature

• So a graph can be plotted using that relation

• It is shown in fig.5.14 below:

Fig.5.14

• We see that, the graph is a straight line

    ♦ $\mathbf\small{\rm{V_t= k_2 \times T_t}}$ will indeed give a straight line

    ♦ It is of the form: y = mx

• After plotting the graph, if we extend it towards the left, it will pass through the origin

    ♦ This is shown by the green dashed line

• The graphs in figs.5.10 and 5.11 also shows the relation between volume and temperature

    ♦ But the difference is this:

          ✰ In figs.5.10 and 5.11, the temperature is plotted in ^oC

          ✰ In fig.5.14, the temperature is plotted in K

• When the temperature is in K, there will not be a y-intercept

    ♦ This is because:

          ✰ At 0 K, the volume of gas is zero

• When the temperature is in ^oC, there will be a y-intercept

    ♦ This is because:

          ✰ At 0 ^oC, the volume of gas is not zero

30. In fig.5.14, all readings were taken while the pressure was constant at p₁

• We can repeat the experiment by applying various temperatures while the pressure is kept constant at another value p₂

• We can repeat the experiment by applying various temperatures while the pressure is kept constant at yet another value p₃

• so on . . .

■ All results can be plotted on the same graph.

    ♦ Such a graph is shown in fig.5.15 below

Fig.5.15

• If extended, all of them will pass through the origin

• Each line in fig.5.15 is called isobar

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• We have completed a discussion on Boyle's law and Charles' law

• We will now see some solved examples based on those two laws. The link is given below:

Solved examples 5.1 to 5.4

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• In the next section, we will see the gaseous state

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Chapter 2.3 - Atomic number and Mass number

In the previous section, we saw the models proposed by J.J. Thomson and Ernst Rutherford. In this section we will see details about atomic number and mass number

1. We saw that in an atom, there are electrons, protons and neutrons
• Each electron has a negative charge of -1.602176 ×10^-19 C 
• Each proton has a positive charge of +1.602176 ×10^-19 C
2. The magnitudes of both the charges are equal
• But the signs are opposite
3. So the number of protons and electrons in an atom must be equal
• Otherwise, the atom will not be electrically neutral
4. The number of protons in an atom is called the atomic number of that atom
• The symbol of atomic number is Z
Some examples:
• The number of protons in hydrogen atom is 1
    ♦ So Z of hydrogen atom is 1
• The number of protons in sodium atom is 11
    ♦ So Z of sodium atom is 11
5. We can write:
Atomic number (Z) of an atom 
= Number of protons in the nucleus of that atom
= Number of electrons in that ‘atom in neutral state’
• ‘Atom in neutral state’ has to be specifically written because, if it is not electrically neutral, the number of electrons will not be equal to the number of protons
6. Next we consider mass number
• Protons and neutrons present in the nucleus are collectively known as nucleons
• The total mass of the nucleons is equal to the mass of the atom
• The total number of nucleons in an atom is called the 'mass number' of that atom
• The symbol of mass number is A
7. We can write:
• Mass number of an atom = Number of protons (Z) + Number of neutrons (n)
• We have seen the details in our previous classes. 
8. We know that every element is represented by a unique symbol
• Z and A can be attached to that symbol
    ♦ Z is written as a subscript on the left side of the symbol 
    ♦ A is written as a superscript on the left side of the symbol 
9. Thus, if ‘X’ is the symbol of an element and Z and A the atomic and mass numbers, we can write: $\mathbf\small{\rm{^A_ZX}}$

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Isobars

• Isobars are elements with same mass number (A) but different atomic numbers (Z)
Example: $\mathbf\small{\rm{^{14}_{\;6}C\;\;\text{and}\;\;^{14}_{\;7}N}}$

Isotopes

Details about isotopes can be written in 6 steps:
1. Consider the atom of hydrogen
• It has 1 proton, 0 neutron and 1 electron
    ♦ So Z = Number of protons = 1
    ♦ Also, A = (Z+n) = (1+0) = 1
• Symbolically we can represent this as $\mathbf\small{\rm{^1_1H}}$
• It is called protium
2. Consider the atom of deuterium
• It has 1 proton, 1 neutron and 1 electron
    ♦ So Z = Number of protons = 1
    ♦ Also, A = (Z+n) = (1+1) = 2
• Symbolically we can represent this as $\mathbf\small{\rm{^2_1D}}$
3. Consider the atom of tritium
• It has 1 proton, 2 neutron and 1 electron
    ♦ So Z = Number of protons = 1
    ♦ Also, A = (Z+n) = (1+2) = 3
• Symbolically we can represent this as $\mathbf\small{\rm{^3_1T}}$
4. In all the above 3 cases, Z is the same. But A are different
• The difference in A is due to the difference in n
■ Elements with same atomic number (Z) but different mass numbers (A) are called isotopes
• $\mathbf\small{\rm{^1_1H,\;\;\,^2_1D\;\;\text{and}\;\;^3_1T}}$ are isotopes
5. If we take a sample of naturally occurring hydrogen, we get the following quantities:
(i) 99.985% of that sample will be $\mathbf\small{\rm{^1_1H}}$
(ii) The rest 0.015% will be $\mathbf\small{\rm{^2_1D}}$ 
(iii) $\mathbf\small{\rm{^3_1T}}$ is found only in traces on earth
6. Let us see some more examples of isotopes:
• Isotopes of carbon:
(i) $\mathbf\small{\rm{^{12}_{\;6}C}}$ has 6 protons and 6 neutrons
(ii) $\mathbf\small{\rm{^{13}_{\;6}C}}$ has 6 protons and 7 neutrons
(iii) $\mathbf\small{\rm{^{14}_{\;6}C}}$ has 6 protons and 8 neutrons
• Isotopes of chorine:
(i) $\mathbf\small{\rm{^{35}_{17}Cl}}$ has 17 protons and 18 neutrons
(ii)$\mathbf\small{\rm{^{37}_{17}Cl}}$ has 17 protons and 20 neutrons

Chemical properties of isotopes

1. The chemical properties of an element are controlled by the number of electrons in it’s atoms
2. The number of neutrons have very little effect on the chemical properties
3. The number of electrons is same as the atomic number (Z)
• So with regard to chemical properties, the number Z is what we consider most
4. All isotopes of an element will have the same Z
• So all isotopes of an element will show the same chemical behaviour

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Now we will see some solved examples:

Solved example 2.1
Calculate the number of protons, neutrons and electrons in $\mathbf\small{\rm{^{80}_{35}Br}}$
Solution:
1. Given : Z = 35, A = 80
2. We have: Number of protons = Number of electrons = Z = 35
3. Also we have:  Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (80-35) = 45

Solved example 2.2
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species
Solution:
1. Given: 
• Number of electrons = 18
• Number of protons = 16
    ♦ So Z = 16
    ♦ From the periodic table, we have: Z = 16 for sulphur (S)
• Number of neutrons = 16
2. We have:  A = Number of neutrons + Number of protons
⇒ A = (16+16) = 32
3. In a neutral atom, the number of electrons will be equal to that of the protons
• But in this case, number of electrons is greater by 2
• So there will be an excess of two negative charges
• So the symbol is: $\mathbf\small{\rm{^{32}_{16}S^{2-}}}$
4. Note:
• If the number of protons and electrons are not equal, the given species is an ion
• We have to determine whether it is a cation or an anion
• If Number of protons > Number of electrons
    ♦ It is a +ve ion (cation)
• If Number of protons < Number of electrons
    ♦ It is a -ve ion (anion)

Solved example 2.3
(i) Calculate the number of electrons which will together weigh one gram
(ii) Calculate the mass and charge of one mole of electrons
Solution:
Part (i)
• Mass of one electron = 9.10939 ×10^-31 kg = 9.10939 ×10^-28 g
• So number of electrons in 1 g 
= $\mathbf\small{\frac{1}{9.10939\times 10^{-28}}=0.1097768\times 10^{28}=1.098\times 10^{27}}$
Part (ii)
• One mole of electrons will contain 6.022 × 10²³ electrons
• Mass of one electron = 9.10939 ×10^-31 kg
• So mass of one mole of electrons = (6.022 × 10²³ × 9.10939 ×10^-31) = 5.48 ×10^-7 kg
• Charge of one electron = -1.602176 ×10^-19 C
• So charge of one mole of electron = (6.022 × 10²³ × -1.602176 ×10^-19) = 9.65 ×10⁴ C

Solved example 2.4
(i) Calculate the total number of electrons present in one mole of methane
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of $\mathbf\small{\rm{^{14}C}}$
(Assume that mass of a neutron = 1.675 × 10^-27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed ?
Solution:
Part (i)
1. The molecular formula of methane is CH₄
• So in one molecule of methane, there are:
    ♦ 1 atom of C
    ♦ 4 atoms of H
2. In one atom of C there are 6 electrons
• In one atom of H there is 1 electron
• So in 1 molecule of methane, there are (1×6 + 4×1) = 10 electrons
3. One mole of methane contains 6.022 × 10²³ molecules of methane
• So in one mole of methane, there will be 6.022 × 10²³ × 10 = 6.022 × 10²⁴  electrons
Part (ii)
1. Molar mass of $\mathbf\small{\rm{^{14}C}}$ is 14 g
⇒ In 14 g of $\mathbf\small{\rm{^{14}C}}$, there will be 6.022 × 10²³ atoms
⇒ In 1 g, there will be (6.022 × 10²³ / 14) atoms
⇒ In 1 mg there will be (6.022 × 10²³ / 14000) atoms 
⇒ In 7 mg there will be (6.022 × 10²³ / 2000) = 3.011 × 10²⁰ atoms
2. Number of neutrons in 1 $\mathbf\small{\rm{^{14}C}}$ atom = (A-Z) = (14-6) = 8
• So total number of neutrons in 7 mg = 8 × 3.011 × 10²⁰ = 24.088 × 10²⁰
3. Mass of this much neutrons = (24.088 × 10²⁰ × 1.675 × 10^-27) 
= 40.35 × 10^-7 
= 4.035 × 10^-6 kg
Part (iii)
1. Molar mass of NH₃ = (14+3) = 17 g
⇒ 17 g of NH₃ at STP will contain 6.022 × 10²³ molecules
⇒ 1 g of NH₃ at STP will contain (6.022 × 10²³ / 17) molecules
⇒ 1 mg of NH₃ at STP will contain (6.022 × 10²³ / 17000) molecules
⇒ 34 mg of NH₃ at STP will contain (6.022 × 2 × 10²⁰) molecules
2. one molecule of NH₃ will contain (7+3) = 10 protons 
• So number of protons in (6.022 × 2 × 10²⁰) molecules 
= 6.022 × 2 × 10 × 10²⁰ 
= 12.044 × 10²¹
= 1.2044 × 10²²
3. Mass of this much protons 
= (1.2044 × 10²² × mass of one proton)
= (1.2044 × 10²² × 1.67262 × 10^-27) 
= 2.015 × 10^-5 kg

Solved example 2.5
How many neutrons and protons are there in the following nuclei ?
$\mathbf\small{\rm{^{13}_{\;6}C,\;\;^{16}_{\;8}O,\;\;^{24}_{12}Mg,\;\;^{56}_{26}Fe,\;\;^{88}_{38}Sr}}$
Solution:
Part (i)
1. Given : Z = 6, A = 13
2. We have: Number of protons = Z = 6
3. We have:  Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (13-6) = 7
Part (ii)
1. Given : Z = 8, A = 16
2. We have: Number of protons = Z = 8
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (16-8) = 8
Part (iii)
1. Given : Z = 12, A = 24
2. We have: Number of protons = Z = 12
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (24-12) = 12
Part (iv)
1. Given : Z = 26, A = 56
2. We have: Number of protons = Z = 26
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (56-26) = 30
Part (v)
1. Given : Z = 38, A = 88
2. We have: Number of protons = Z = 38
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (88-38) = 50

Solved example 2.6

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35
(ii) Z = 92 , A = 233
(iii) Z = 4 , A = 9
Solution:
Part (i)
• Given: Z = 17, A = 35
    ♦ From the periodic table, we have: Z = 17 for chlorine (Cl)
• So symbol is $\mathbf\small{\rm{^{35}_{17}Cl}}$
Part (ii)
• Given: Z = 92, A = 233
    ♦ From the periodic table, we have: Z = 92 for uranium (U)
• So symbol is $\mathbf\small{\rm{^{233}_{92}U}}$
Part (iii)
• Given: Z = 4, A = 9
    ♦ From the periodic table, we have: Z = 4 for beryllium (Be)
• So symbol is $\mathbf\small{\rm{^{9}_{4}Be}}$

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In the next section we will see some more solved examples. We will also see the drawbacks of Rutherford's model

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