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 Vt, we can write:
The increase in volume = (Vt-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
♦ Vt and t are variables
• So we plot a Vt 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:
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| 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
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 Vt 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
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| 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 Vt = 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)
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| 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: Tt 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:
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| 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
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| Fig.5.15 |
• If extended, all of them will pass through the origin
• Each line in fig.5.15 is called isobar
• 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:
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