In the previous section we completed a discussion on thermal properties of matter. In this chapter, we will see thermodynamics.
First we will see the difference between macro properties and micro properties. It can be written in 4 steps:
1. Consider the cylinder and piston shown in fig.12.1(a) below:
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| Fig.12.1 |
• The cylinder contains some gas. The molecules of the gas are shown in magenta color.
2. We know that, the gas in the cylinder will have:
♦ a certain volume, V
♦ a certain pressure, P
♦ a certain temperature, T
■ The above three properties are called macro properties of the gas.
♦ They are related to the total mass of the gas.
✰ All portions of the mass will be experiencing the same pressure.
✰ All portions of the mass will be experiencing the same temperature.
• We will see more such macro properties in later sections.
3. Consider the individual molecules of the gas. Let us write some properties of those molecules:
(i) Each molecule will have it’s own speed and direction.
• In other words, each molecule will have it’s own velocity.
• Some molecules will be moving towards left, some others towards right, some others upwards so on . . .
• Let us number the molecules as 1, 2, 3, . . . , n
♦ Then we can write the velocities as: v1, v2, v3, . . . , vn
(ii) Since, each molecule has it’s own velocity, each of them will be having it’s own kinetic energy also.
♦ We can write the kinetic energies as: KE1, KE2, KE3, . . . , KEn
(iii) Since the molecules are always in motion, they will be striking against the walls of the cylinder.
• While striking the walls, each molecule will be exerting a force on the walls.
♦ We can write the forces as: F1, F2, F3, . . . , Fn
■ The above three properties are called micro properties of the gas.
♦ We are considering such properties at the micro level.
4. The macro properties and micro properties are related to each other.
• But in thermodynamics, we do not need the micro properties. We need the macro properties only.
• We can arrive at the required results with out knowing the micro properties.
• In fact, the discoveries in thermodynamics were made long before individual particles like molecules, atoms, electrons, protons etc., were discovered.
• While writing the steps, we will learn about some of the common ‘thermodynamic terms’ also.
1. Let us place some weights on the piston in fig.12.1(a)
• Due to the weights, the piston will move downwards and compress the gas. This is shown in fig. 12.1(b).
• Let PA and VA be the pressure and volume of the compressed gas.
2. So we have a pressure PA and the corresponding volume VA
• We can mark them in a P-V Diagram.
♦ It is shown in fig.12.2(a) below:
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| Fig.12.2 |
• A P-V Diagram is a graph with,
♦ pressure plotted along the y-axis.
♦ volume plotted along the x-axis.
3. We know that, in a graph, the points that we plot will be in the coordinates form (x,y).
• So in our present case, we have the coordinates (VA, PA).
♦ We will call it: Point A.
4. Point A (VA,PA) is an equilibrium point.
• That means, at point A, the gas is in equilibrium with it’s surroundings.
♦ It’s pressure remains constant at PA
♦ It’s volume remains constant at VA
• We can measure the temperature TA also.
♦ It will remain constant at TA
5. Next, we remove some weights from the top of the piston.
• The gas now feels a lesser compressive force. It will expand.
• Thus the piston will move upwards.
♦ The final position of the piston is shown in fig.12.1(c).
6. Let PB and VB be the pressure and volume of the expanded gas.
• Thus we get a new point B(VB, PB)
♦ This point is also plotted in the fig.12.2(a)
7. The following facts may be noted:
• In the expanded condition, volume will be larger.
♦ So in the P-V Diagram, VB will be on the right side of VA
• In the expanded condition, pressure will be lesser.
♦ So in the P-V Diagram, PB will be at the bottom side of PA
■ So in effect, point B will be to the bottom-right of point A.
8. Point B (VB, PB) is also an equilibrium point.
• That means, at point B, the gas is in equilibrium with it’s surroundings.
♦ It’s pressure remains constant at PB
♦ It’s volume remains constant at VB
• We can measure the temperature TB also.
♦ It will remain constant at TB
• We have complete details about each of those two states.
• But we want the details about intermediate states also.
10. What do we mean by ‘intermediate states’?
An explanation can be written in 5 steps:
(i) We have two pressure values: Initial pressure PA and final pressure PB
(ii) PB is lesser than PA
• So, while traveling from A to B, the gas would have experienced a series of pressure values.
♦ Each value in that series would be lesser than it’s previous value.
(iii) We want every values in that series.
♦ We want the corresponding volume values also.
(iv) If we can obtain the pressure values and the corresponding volume values, we will be able to write the coordinate points:
(P1, V1), (P2, V2), (P3, V3) . . .
(v) These are the intermediate states. They will lie in the path between A and B.
This is shown in fig.12.2(b).
■ Let us make an attempt to find those intermediate states. The following steps from (11) to (14) gives a description about our attempt:
11. First, we were at state A. We saw it in fig.12.1(b)
• There, we removed some weights.
• When the weights are removed, the piston rises up to reach state B. We saw it in fig.(c).
12. This is a sudden change. As soon as we remove the weights, the piston will rise up.
• The piston will shoot up beyond the position in fig.(c)
♦ This is shown in fig.(d).
• From this top most point, the piston will fall downwards.
♦ It will fall to a point below the position in fig.(c)
♦ This is shown in fig.(e)
13. The piston will oscillate vertically between the two extreme points.
♦ The top extreme point is the one in fig.(d).
♦ The bottom extreme point is the one in fig.(e).
• The oscillation will continue for some time.
• Finally, it will come to rest at the position in fig.(c).
14. We want the P, V and T values, beginning from the instant when the weights are removed.
■ Here arises a problem:
Due to the oscillation, the P, V and T values will be continuously fluctuating. This can be explained in 4 steps:
(i) During the oscillation, the piston will be continuously moving upwards and downwards.
(ii) Consider one upward motion of the piston.
• During this upward motion, the pressure near the piston will be lesser than the pressure away from the piston.
• That means:
♦ The upper portion of the gas mass will be having a lesser pressure.
♦ The lower portion of the gas mass will be having a greater pressure.
• There will be corresponding variations in temperature also:
The upper portion of the gas mass will be having a different temperature from the lower portion.
• We can record the P and T values only if the whole mass have uniform values.
• That means, to record P and T:
♦ Pressure must be uniform through out the whole gas mass.
♦ Temperature must be uniform through out the whole gas mass.
(iii) Now consider one downward motion of the piston.
• During this downward motion, the pressure near the piston will be higher than the pressure away from the piston.
• That means:
♦ The upper portion of the gas mass will be having a greater pressure.
♦ The lower portion of the gas mass will be having a lesser pressure.
• There will be corresponding variations in temperature also:
The upper portion of the gas mass will be having a different temperature from the lower portion.
• We can record the P and T values only if the whole mass have uniform values.
• That means, to record P and T:
♦ Pressure must be uniform through out the whole gas mass.
♦ Temperature must be uniform through out the whole gas mass.
(iv) Thus it is clear that, due to the oscillation, it is impossible to record the P, V and T values.
■ So we adopt another procedure. The following steps from (15) to (20) gives a description about that procedure:
15. Fig.12.3(a) below shows the initial state.
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| Fig.12.3 |
• Earlier, we removed four rows of weights from each vessel.
♦ All those weights were removed together.
• This time, we remove the weights in stages.
♦ First we remove only the top most row from the vessels.
♦ So the piston will rise only a little.
♦ This is shown in fig.12.3(b).
• Since there is only a tiny rise, there will not be much oscillation.
♦ We will be able to measure P, V and T.
♦ We will call them P1, V1 and T1
16. Next, we remove the second row of weights.
♦ The piston will rise only a little.
♦ This is shown in fig.12.3(c).
• Since there is only a tiny rise, there will not be much oscillation.
♦ We will be able to measure P, V and T.
♦ We will call them P2, V2 and T2
17. Next, we remove the third row of weights.
♦ The piston will rise only a little.
♦ This is shown in fig.12.3(d).
• Since there is only a tiny rise, there will not be much oscillation.
♦ We will be able to measure P, V and T.
♦ We will call them P3, V3 and T3
18. We continue like this and reach the final point B.
19. Theoretically, we must remove only an infinitesimal weight in each stage.
• Infinitesimal means very small
♦ It is a small quantity which is very close to zero.
♦ So we must remove only about 0.00001 grams at a time.
• So in each stage, the piston will rise only an infinitesimal height.
♦ This will ensure that, the gas is in equilibrium all the time.
• We will get a large number of points between A and B.
♦ So we will get a smooth curve between A and B.
♦ This curve is shown in red color fig.12.4(a) below:
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| Fig.12.4 |
20. When we do the experiment in this way, the process is almost static.
■ So we call it: A quasi static process
♦ ‘Quasi’ means ‘almost same as’.
♦ ‘Quasi static’ means 'almost static'.
• Note that, the process cannot be perfectly static because, then there would be the first state A only
Now we can discuss about reversible process. It can be explained in 8 steps:
1. Imagine that during the above process, we remove an infinitesimal weight.
• The piston will move up an infinitesimal height.
2. During the upward motion, the gas does some work in pushing the piston.
• But the edges of the piston encounters friction with the sides of the cylinder.
• So some work done by the gas is lost as heat (frictional heat).
3. The ‘removal of a certain weight’ should enable the piston to ‘rise by a certain height’.
• We can calculate this ‘certain height’ theoretically.
■ But the actual height achieved will be always less (than the theoretical value) because of the friction.
4. Next, we put the weight back on.
• The piston moves downwards.
• During this downward motion also, there will be friction between the piston and the sides.
• So here also, some work is lost as heat.
5. The ‘addition of the weight’ should ‘push the piston by a certain depth’.
• We can calculate this ‘certain depth’ theoretically.
■ But the actual depth achieved will be always less (than the theoretical value) because of the friction
6. If there is no friction, we will attain the ideal condition:
♦ Remove the weight. The piston rises by the theoretical height.
♦ Put the weight back on. The piston lowers to the original level.
■ Such an ideal process is called reversible process.
In a reversible process, we can 'reach back to a point along the same path' by 'undoing an action'.
♦ This is shown in fig.12.4(b) above.
7. So ‘ideal condition’ with no friction, no viscosity and no air resistance, is a requirement for a process to be reversible.
• There is one more requirement. It can be explained in 3 steps:
(i) If at any stage, we remove a large weight from fig.12.3(a) above, the pressure and temperature will suffer a large change.
(ii) After such a large change, we will not be able to get back to the previous state by putting back the weight.
(iii) So the second requirement for the process to be reversible is that:
Only infinitesimal weights are removed at each stage.
8. All processes which are not reversible are called irreversible processes
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