Chapter 7.10 - Effect of Pressure and Temperature on Equilibrium

In the previous section, we saw how change in concentration affects equilibrium. In this section, we will see three more factors which  affect equilibrium

Factor 2: Pressure
• This can be written in 8 steps:
1. Changes in pressure affects equilibrium of reactions in which at least one reactant or product is a gas
2. Suppose that, the system is inside a cylinder fitted with a piston
• Let P₁ and V₁ be the pressure and volume at equilibrium
• If we push down the piston to make the volume $\mathbf\small{\rm{\frac{V_1}{2}}}$, what will be the change in pressure?
3. We know that PV = a constant
• So P₁V₁ = P₂V₂
⇒ $\mathbf\small{\rm{p_1V_1=p_2 \times \frac{V_1}{2}}}$
⇒ P₂ = 2P₁
• That means, pressure becomes double the initial value
4. We know that:
• P₁ and P₂ are the total pressures
    ♦ Total pressure of the system is the sum of the partial pressures
    ♦ Partial pressure of a gas depends on it’s number of moles
• Now, total pressure has increased to a new value P₂
    ♦ So partial pressures will also increase
5. According to Le Chatelier’s principle, the system will try to reduce the effect of this increased pressure
• That means, the pressure must decrease from P₂
• For that, the reaction will proceed in that direction which will result in a lesser number of moles
6. An example:
• In the reaction CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), there are four gaseous moles on the reactant side, but only two gaseous moles in the product side
• So, when the pressure is increased, the reaction moves in the forward direction
7. Another example:
• In the reaction C(s) + CO₂(g) ⇌ 2CO(g), there is only one gaseous mole on the reactant side, but two gaseous moles in the product side
• So, when the pressure is increased, the reaction moves in the backward direction
8. Note that, with regard to 'change in pressure':
• We need not consider solids and liquids because, a change in pressure will not affect their volumes
• If volumes are not affected, concentrations are also not affected

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Effect of adding an inert gas

• This can be written in 2 steps:
1. Consider a system in equilibrium
• If an inert gas is added into the system, it will not take part in the reaction
• So the concentrations of reactants and products will not change
2. Thus we can write:
In a system at equilibrium, if we add an inert gas with out changing the pressure, the equilibrium will not be affected

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Factor 3: Temperature
• Recall the two factors that we have already seen:
(i) When concentrations are changed, the K_c does not change. The concentrations re-adjust to bring back K_c to the original value
(ii) When pressure is changed, the K_c(or K_p) does not change. The concentrations re-adjust to bring back K_c(or K_p) to the original value

• But the third factor is different. When temperature changes, K_c(or K_p) attains a new value
• This can be explained using an example. It can be written in 5 steps:
1. Consider the reaction: H₂(g) + I₂(g) ⇌ 2HI(g); ΔH = -10.4 kJ mol^-1
• It is clear that, the forward reaction is exothermic. 10.4 kJ mol^-1 will be released
2. At 500 K, the Kp of this reaction is: 160
• If the temperature is increased to 700 K, the K_p becomes 54
◼ We see that, when temperature is increased, the equilibrium constant decreases
the following steps from (3) to (5) will give the reason:
3. According to Le Chatelier’s principle, the reaction will move in that direction which reduces the effect
• In our present case, temperature is increased. So the system tries to reduce the temperature
4. For that, the reaction will move in that direction which absorbs heat
    ♦ The forward reaction in (1) is exothermic which releases heat
    ♦ The backward reaction is endothermic which absorbs heat
• So the backward reaction is preferred
• So more reactants will be produced and more products will be consumed
5. That means, numerator (in the expression $\mathbf\small{\rm{K_p=\frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b}}}$) decreases
• Thus the equilibrium constant decreases

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Factor 4: Catalyst
This can be written in 4 steps:
1. A catalyst is added to speed up the reaction
2. But it speeds up both forward and backward reactions
• Thus the equilibrium is attained in a smaller time
3. Since the rates of both forward and backward reactions are increased, the concentrations of reactants and products remain the same
4. So the value of equilibrium constant does not change

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Now we will see some solved examples:
Solved example 7.36
Does the number of moles of reaction products increase, decrease or remain
same when each of the following equilibria is subjected to a decrease in pressure
by increasing the volume?
(a) PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)
(b) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
(c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Solution:
Part (a):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure decreases, the system will try to increase the number of gaseous moles
    ♦ So the reaction will proceed in the forward direction
    ♦ Thus the number of moles of reaction products will increase
Part (b):
• Total number of gaseous moles in the reactant side = 2
• Total number of gaseous moles in the product side = 1
◼ When pressure decreases, the system will try to increase the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
    ♦ Thus the number of moles of reaction products will decrease
Part (c):
• Total number of gaseous moles in the reactant side = 4
• Total number of gaseous moles in the product side = 4
◼ When pressure decreases, the system will try to increase the number of gaseous moles. But in this case, the number of gaseous moles is same for both reactants and products
    ♦ So the decrease in pressure has no effect in the direction
    ♦ Thus the number of moles of reaction products will remain the same

Solved example 7.37
Which of the following reactions will get affected by increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or
backward direction.
(i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(s) ⇌ 2CO(g)
(iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Solution:
Part (i):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
Part (ii):
• Total number of gaseous moles in the reactant side = 3
• Total number of gaseous moles in the product side = 3
◼ When pressure increases, the system will try to decrease the number of gaseous moles. But in this case, the number of gaseous moles is same for both reactants and products
    ♦ So the increase in pressure has no effect in the direction
Part (iii):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
Part (iv):
• Total number of gaseous moles in the reactant side = 3
• Total number of gaseous moles in the product side = 1
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the forward direction
Part (v):
• Total number of gaseous moles in the reactant side = 0
• Total number of gaseous moles in the product side = 1
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
Part (vi):
• Total number of gaseous moles in the reactant side = 9
• Total number of gaseous moles in the product side = 10
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction

Solved example 7.38
The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024K
H₂ (g) + Br₂ (g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a
sealed container at 1024K.
Solution:
1. At the beginning of the reaction, only HBr is present
• So we have to write the given equation in reverse order:
2HBr(g)  ⇌ H₂ (g) + Br₂ (g)
• The equilibrium constant of this reaction will be: $\mathbf\small{\rm{\frac{1}{1.6 \times 10^5}}}$
2. At equilibrium, let x be the pressure of H₂ and Br₂
• Then the pressure of HBr will be (10.0 - 2x)
3. Thus the equilibrium constant will be given by: $\mathbf\small{\rm{\frac{x^2}{(10.0-2x)^2}}}$
• So we can write: $\mathbf\small{\rm{\frac{x^2}{(10.0-2x)^2}=\frac{1}{1.6 \times 10^5}}}$
4. Taking reciprocals, we get: $\mathbf\small{\rm{\frac{(10.0-2x)^2}{x^2}=\frac{1.6 \times 10^5}{1}}}$
⇒ $\mathbf\small{\rm{\frac{(10.0-2x)}{x}=\sqrt{\frac{16 \times 10^4}{1}}=400}}$
• Thus we get x = 0.0248
5. So we can write:
• At equilibrium,
   ♦ p_H2 = p_Br2 = x = 0.0248 bar
   ♦ p_HBr = (10 - 2x) = (10 - 0.0496) = 9.95 bar

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Link to a few more solved examples are given below:

Solved examples 7.39 to 7.45

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In the next section, we will see the ionic equilibrium in solution


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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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