In the previous section we saw the first law of thermodynamics. In this section, we will see work done during a thermodynamic process.
The method of calculating the work can be explained in 18 steps:
1. We have seen the gas inside the cylinder in fig.12.5 at the beginning of the previous section.
• Suppose that, the gas is expanding.
♦ The expanding gas will push the piston upwards.
♦ That means, the gas is doing work on the piston.
2. By a suitable mechanism, the piston can be connected to the wheel of an automobile.
• Then the upward motion of the piston will rotate the wheel.
• So in effect, we are using the ‘expansion of the gas’ to rotate the wheel.
• That means, the work done by the gas is converted into kinetic energy of the automobile.
3. We want to calculate the amount of work done by the gas.
• In fig.12.7(b) below, the green arrows indicate the pressure exerted by the gas on the piston.
♦ Let this pressure be P.
 |
| Fig.12.7 |
• The piston will be having a certain ‘cross sectional area’ A.
♦ So the pressure P is acting on an area A.
• We know that: Pressure = Force⁄Area
♦ So Force = Pressure × Area
♦ So in our present case, force exerted by the gas = P × A
4. As a result of this force, let the piston move up a distance of Δd.
♦ This is shown in fig.12.7(c)
• We know that: Work done = Force × distance
• So the work done by the gas in moving the piston by a distance of Δd.
= (P × A) × Δd
• We can rearrange this as:
Work done = P × (A × Δd)
5. But (A × Δd) will be the volume of a prism (in our present case, the prism is a cylinder) whose base area is A and height is Δd.
♦ This is shown in fig.12.7(d)
• In our present case, it is the volume between the initial and final positions of the piston.
• So (A × Δd) is the change in volume ΔV
6. Substituting for '(A × Δd)' in (4), we get:
Eq.12.2: Work done by the gas in moving the piston by a distance of Δd = P × ΔV
• So we can find the work done by multiplying pressure and change in volume.
7. But there is a problem. It can be explained in 3 steps:
(i) Consider fig.a. It shows the final position of the piston.
♦ For better comparison, it is drawn to the left of the initial state (fig.b)
• The piston has moved through a large distance d.
(ii) In such a case, volume has increased by a large amount.
• So pressure will have changed.
♦ The pressure in fig.a will be different from the pressure in fig.b.
• If the piston is at some intermediate point, then also the pressure will be different from the pressure in fig.b.
• It is clear that, P varies continuously as the thermodynamic process proceeds.
(iii) We want to use the formula: Work done = P × ΔV
• But if P is continuously changing, we cannot input a value for P.
■ So we adopt a new procedure. The following steps from (8) to (14) explains this new procedure:
8. Fig.12.8(a) below shows a P-V Diagram.
♦ A (VA,PA) is the initial point.
♦ B (VB, PB) is the final point.
• It is clear that, pressure varies with volume.
(If pressure was a constant, the graph would have been a straight horizontal line)
• A number of points (yellow dots) are marked on the graph.
 |
| Fig.12.8 |
9. Next, we draw some rectangles based on those dots.
• The rectangles must satisfy an important condition:
♦ The dots must be situated at the top-left corner of each rectangle.
• When the rectangles are drawn in this way, we get:
♦ The heights of the rectangles are equal to the pressure values at the dots.
10. The bases of the rectangles are also important.
• The bases of the rectangles are on the volume axis. The widths of those bases will give the changes in volume.
• For example:
♦ At the first yellow dot, pressure is P1 and volume is V1
♦ At the second yellow dot, pressure is P2 and volume is V2
♦ So the width of the base of the second rectangle = V2 – V1
♦ This is the change in volume ΔV
✰ This is the change when,
✰ the gas moves from the first yellow dot to the second yellow dot.
11. Thus we can obtain the heights and widths of all rectangles.
• Now, when the gas moves from one dot to the next, we assume that the pressure remains constant.
• For example:
♦ When the gas moves from the first dot to the second, the pressure surely changes from P1 to P2
♦ But since the width of the rectangle is small, we assume that: Pressure remains constant at P1
♦ So for the volume change (ΔV = V2 – V1), the pressure is P1
12. Since the pressure is constant at P1, we can use the equation 12.2
• We get: work done while moving the piston from V1 to V2 = P1 ΔV = P1 (V2 – V1)
• Clearly, 'P1 (V2 – V1)' is the area of the second rectangle.
13. So it is clear that:
• Work done when the volume changes from V1 to V2 is equal to the area of the second rectangle.
[Let us do a dimensional check:
• We want to check whether
♦ The dimensions of (P ΔV)
♦ is same as
♦ The dimensions of work
• The dimensions of (P ΔV) are:
$\mathbf\small{\rm{\frac{MLT^{-2}}{L^2} \times L^3\;=\;ML^2T^{-2}}}$
• The dimensions of work are:
$\mathbf\small{\rm{MLT^{-2}\times L\;=\;ML^2 T^{-2}}}$
• Both dimensions are the same]
14. The total work:
• Since we know the heights and widths of all the rectangles, we can calculate the areas of all the rectangles.
• The sum of all those areas will give the total work done when the piston moves from A to B.
• Thus we succeed in calculating the work done.
15. But now another problem arises. It can be explained in 2 steps:
(i) In fig.12.8 (a), the top right corners of the rectangles are above the curve.
♦ So a small triangular portion of each rectangle is outside the curve.
(ii) Thus some 'unwanted areas' are included while calculating the total work.
♦ This will lead to error.
16. To solve this problem, we reduce the widths of the rectangles.
• Then the triangular portions will become smaller.
• As the widths decrease, sizes of the unwanted triangles also decrease.
■ If we make the widths infinitesimal, there will not be any triangular portions projecting out.
• The total area calculated by such a method will give the accurate amount of work.
17. But when the widths are infinitesimal, the number of rectangles will be very large.
♦ A few such rectangles are shown in fig.b
• We will have to calculate the areas of all rectangles from point A to point B.
• Here, calculus comes to our help. Using calculus, we can easily find the area below the curve.
(At this stage, we do not have to learn the mathematical details of calculus. All we need is the final expression for determining the work. We will see such an expression later in this section)
18. Thus the above 17 steps help us to arrive at the fact that:
The area below the P-V Diagram will give the work done by the gas or the work done on the gas.
• The gas inside the cylinder can be subjected to four types of thermodynamic processes. They are:
(i) Isothermal process
(ii) Isobaric process
(iii) Isochoric process
(iv) Adiabatic process
We will now see the basics of each of the above four processes.
Isothermal process
This can be explained in 7 steps:
1. The word 'iso' means equal.
• So isothermal process is a process in which the temperature remains the same from start to finish.
2. Consider the gas inside the cylinder.
• Let us compress the gas using the piston.
♦ So work is done on the gas. It's internal energy will increase.
• Due to the increase in internal energy, it's temperature will also increase.
3. For the process to be isothermal, there must be no increase in temperature.
• So we put the cylinder in contact with a thermal reservoir.
• A thermal reservoir is a body of large mass.
♦ It also has a large specific heat capacity.
• So it can remain at the same temperature for a long time.
♦ A lake of water is a good example for a thermal reservoir.
■ When the cylinder is kept in contact with a cold reservoir, the heat will flow out to the reservoir, and so the temperature will remain constant.
4. Similarly, when work is done by the gas, it’s internal energy will decrease and so the temperature will decrease.
• To keep the temperature constant, we put the cylinder in contact with a hot reservoir.
• So heat will flow into the cylinder and temperature will remain the same.
5. We know that, if there is no temperature change, there will not be any change in internal energy. That means, ΔU = 0
• So applying Eq.12.1, (when work is done on the gas) we get:
0 = Q + W
⇒ W = -Q
• That means, all the work done on the gas will be released as 'heat from the gas'.
6. Applying Eq.12.1 (when work is done by the gas) we get:
0 = Q – W
⇒ W = Q
• That means, all the heat supplied to the gas is converted into 'work done by the gas'.
7. Let us calculate the work done on or by the gas. It can be written in 9 steps:
(i) Fig.12.9(a) below shows the P-V Diagram of an isothermal process.
• The red curve indicates that, the gas experiences various pressures and volumes.
• But all the while, the temperature will remain constant.
• So in this case, the red curve is called an isotherm.
 |
| Fig.12.9 |
(ii) In fig.(b), vertical dashed lines are drawn through VA and VB
(iii) So we have a region with well defined boundaries.
• This region is shaded with magenta hatch lines.
♦ The vertical dashed lines through VA and VB form the left and right boundaries.
♦ The x-axis forms the bottom boundary.
♦ The red curve forms the top boundary.
(iv) So three boundaries are horizontal/vertical straight lines. Only the top boundary is curve shaped.
• If the top boundary was also a straight line, we could have calculated the area very easily.
(v) However, we need not worry much because, the curve is well defined.
• The curve is a well defined function.
• In simple terms, we can say: Our curve has a unique equation.
• This is just like a line having a unique equation: y = mx + c.
(vi) Let us find the equation of our curve:
♦ We know the ideal gas equation: PV = nRT.
♦ If there is no leakage of gas, n will be a constant.
♦ R is the universal gas constant.
♦ Since our present process is isothermal, T is also a constant.
(vii) So in effect, the equation of the red curve in fig.12.9 is: PV = A constant.
♦ This is same as: $\mathbf\small{\rm{P=\frac{A \, Constant}{V}}}$
♦ It is of the form: $\mathbf\small{\rm{y=\frac{A \, Constant}{x}}}$
• In math classes, we see the graph of $\mathbf\small{\rm{y=\frac{A \, Constant}{x}}}$
♦ That graph has a shape similar to the red curve in fig.12.9
(viii) Also recall that, $\mathbf\small{\rm{P=\frac{A \, Constant}{V}}}$ is the relation given by Boyle’s law
• Robert Boyle did experiments on the gas while keeping n and T constant.
(ix) Now, since the top boundary is well defined, calculus will do the mathematical works.
• The result is:
Area of the magenta hatched region = $\mathbf\small{\rm{n R T\,\,ln \frac{V_B}{V_A}}}$
So we get Eq.12.3:
Work done in an isothermal process = $\mathbf\small{\rm{n R T\,\,ln \frac{V_B}{V_A}}}$
PREVIOUS
CONTENTS
NEXT
Copyright©2020 Higher Secondary Physics. blogspot.in - All Rights Reserved
No comments:
Post a Comment