In the previous section we saw the details about adiabatic process. In this section, we will see cyclic process
While learning about cyclic processes, we will be using a lot of P-V Diagrams. Also, we will need to carefully record internal energy changes. So we will first see how a P-V Diagram is related to the internal energy of the gas. Some basics can be written in 5 steps:
1. Consider any one point in the P-V Diagram. Let it be (V1, P1)
• So, at that point, gas has a definite volume V1 and a definite pressure P1
2. In addition to definite volume and pressure at that point, the gas will have:
♦ a definite temperature T1
♦ a definite internal energy U1
3. So we can denote that point as (V1, P1, T1, U1)
4. When the gas is at another point 2, the values will be: (V2, P2, T2, U2)
• This is shown in fig.12.12(a) below
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| Fig.12.12 |
5. Consider the last terms: U1 and U2
• Initially the internal energy is U1
• In that condition,
♦ We add heat Q to the gas or remove heat Q from the gas
♦ We do some work W on the gas or the gas does a work W
• After the above Q and W interactions, U1 becomes U2
Let us see a solved example which will demonstrate the above facts
Solved example 12.7
A gas is initially at point 1. It’s coordinates are: (V1, P1, T1, U1)
(a) When at point 1, a heat of 80 J is taken away from it. Also a work of 120 J is done on it. What is the ΔU experienced by the gas ?
(b) When at point 1, a heat of 90 J is added to the gas. What is the work done on/by gas? What is the ΔU experienced by the gas ?
(c) When at point 1, the gas does a work of 40 J. What is the heat added/taken? What is the ΔU experienced by the gas ?
(d) When at point 1, a work of 40 J is done on the gas. What is the heat added/taken? What is the ΔU experienced by the gas ?
Solution:
Part (a):
• We have: U2 = U1 -80 +120
• So ΔU = (U2 - U1) = 40 J
Part (b):
(i) Let us assume that, work is done on the gas
• We get: U2 = U1 +90 +W
⇒ ΔU = (U2 - U1) = (90 + W)
(ii) (V1, P1, T1, U1) and (V2, P2, T2, U2) are fixed values
⇒ U1 and U2 are fixed values
⇒ (U2 - U1) has a fixed value
⇒ ΔU has a fixed value
(iii) From part (a), we obtained ΔU as 40 J
• So the result in (i) becomes: 40 = 90 + W
• Thus we get: W = -50
(iv) This is a negative value. So our assumption (that work is done on the gas) in (i) is wrong
• We must write: U2 = U1 + 90 - W
• From this we get: 40 = ΔU = (U2 - U1) = (90 - W)
⇒ W = (90 - 40) = 50 J
• So the final answers are:
♦ ΔU = 40 J
♦ Gas does a work of 50 J
Part (c):
(i) Let us assume that, heat is added to the gas
• We get: U2 = U1 +Q -40
⇒ ΔU = (U2 - U1) = (Q - 40)
(ii) (V1, P1, T1, U1) and (V2, P2, T2, U2) are fixed values
⇒ U1 and U2 are fixed values
⇒ (U2 - U1) has a fixed value
⇒ ΔU has a fixed value
(iii) From part (a), we obtained ΔU as 40 J
• So the result in (i) becomes: 40 = Q - 40
• Thus we get: Q = 80
• So the final answers are:
♦ ΔU = 40 J
♦ Gas receives a heat of 80 J
Part (d):
(i) Let us assume that, heat is added to the gas
• We get: U2 = U1 +Q +40
⇒ ΔU = (U2 - U1) = (Q + 40)
(ii) (V1, P1, T1, U1) and (V2, P2, T2, U2) are fixed values
⇒ U1 and U2 are fixed values
⇒ (U2 - U1) has a fixed value
⇒ ΔU has a fixed value
(iii) From part (a), we obtained ΔU as 40 J
• So the result in (i) becomes: 40 = Q +40
• Thus we get: Q = 0
• So the final answers are:
♦ ΔU = 40 J
♦ No heat is added/taken
Note: In parts (b), (c) and (d) of this solved example, we were able to determine two unknown quantities. This is possible because, U1 and U2 are fixed quantities
1. In fig.12.12(b) above,
♦ The gas first travels from point 1 to point 2 along the red path
♦ Then it travels from point 2 back to point 1 along the green path
2. When it reach back at point 1, it’s internal energy will be U1
■ So the net change in internal energy ΔU = U1 – U1 = 0
The following two solved examples will demonstrate the application of this fact
Solved example 12.8
A gas is initially at point 1 (V1, P1, T1, U1)
(a) When at point 1, a heat of 400 J is given to the gas. Also a work of 100 J is done by the gas. When these Q and W are effected, the gas reaches point 2 (V2, P2, T2, U2). What is the ΔU experienced by the gas while moving from 1 to 2
(b) From point 2, the gas is taken back to point 1. The Q and W during this process are not known. What is the ΔU experienced by the gas while moving from 2 to 1
(c) When at point 1, a work of 400 J is done on the gas. What is the heat Q added/taken? What is the ΔU experienced by the gas? Assume that, as a result of this '400 J work' and 'heat Q', the gas reaches point 2
Solution:
Part (a):
• We have: U2 = U1 +400 -100 = U1 + 300
• So ΔU = (U2 - U1) = 300 J
Part (b):
1. Let the unknown heat and work be Q and W
2. We have: U1 = U2 ±Q ±W
♦ '±' is given before Q because, we do not know whether heat is added or removed
♦ '±' is given before W because, we do not know whether work is done on the gas or by the gas
3. ΔU = (U1 - U2) = (±Q ±W)
• We cannot find ΔU using '(±Q ±W)' because, neither Q nor W is known
• But ΔU is equal to '(U1 - U2)'. The values of U1 and U2 will not change
♦ We have already calculated (U2 - U1) in part (a)
• Thus we get: ΔU = (U1 - U2) = -(U2 - U1) = -300 J
Part (c):
(i) Let us assume that, heat is added to the gas
• We get: U2 = U1 +Q +400
⇒ ΔU = (U2 - U1) = (Q + 400)
(ii) We have already calculated '(U2 - U1)' as 300 J
• So the result in (i) becomes: 300 = Q +400
• Thus we get: Q = -100 J
(iii) This is a negative value. So our assumption (that heat is added to the gas) in (i) is wrong
• We must write: U2 = U1 -Q +400
• From this we get: 300 = ΔU = (U2 - U1) = (-Q +400)
⇒ Q = (400 - 300) = 100 J
• So the final answers are:
♦ ΔU = 300 J
♦ Gas releases a heat of 100 J
Solved example 12.9
A gas is initially at point 1 (V1, P1, T1, U1). It expands and reaches point 2 (V2, P2, T2, U2). The path was along the red curve. See fig.12.13 below. To reach point 2 from point 1, an alternate path 1-3-2 is used. When this path is used, a heat of 515 J is to be added to the gas. How much heat will be required if the path 1-4-2 is used ? Given that:
P1 = 2.10 × 105 N/m2, P2 = 1.05 × 105 N/m2, V1 = 2.25 × 10-3 m3 and V2 = 4.5 × 10-3 m3
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| Fig.12.13 |
Solution:
1. Consider the path 1-3-2
• We have: U2 = U1 + 515 ±W
2. While moving from 1 to 3, volume expands. So work is done by the gas
• Also, while moving from 3 to 2, volume remains the same. So no work is done
• Thus, in effect, in the path 1-3-2, a work of W is done by the gas
• So the sign of W in (1) is negative
• We get: U2 = U1 + 515 - W
3. Our next task is to find W
• The path 1-3 indicates an isobaric process
• So work done = P ΔV = P1 (V2 - V1)
= 2.10 × 105 × (4.5 - 2.25) × 10-3 = 472.5 J
4. Substituting this W in (2), we get:
ΔU = (U2 - U1) = (515 - 472.5) = 42.5 J
5. Consider the path 1-4-2
• We have: U2 = U1 ±Q ±W
6. While moving from 4 to 2, volume expands. So work is done by the gas
• Also, while moving from 1 to 4, volume remains the same. So no work is done
• Thus, in effect, in the path 1-4-2, a work of W is done by the gas
• So the sign of W in (5) is negative
• We get: U2 = U1 ± Q - W
7. Our next task is to find W
• The path 4-2 indicates an isobaric process
• So work done = P ΔV = P2 (V2 - V1)
= 1.05 × 105 × (4.5 - 2.25) × 10-3 = 236.25 J
8. Substituting this W in (6), we get:
ΔU = U2 - U1 = (± Q - 236.25)
9. But we already know ΔU from (4)
So we get: 42.5 = (± Q - 236.25)
10. Let us assume that, heat is added
• Then the result in (9) becomes:
42.5 = Q - 236.25
⇒ Q = (42.5 + 236.25) = 278.75 J
11. This is a positive value. So the assumption made in (10) is correct
• The final answer is:
When the gas moves along the path 1-4-2, a heat of 278.75 J should be added to the gas
Now we will see cyclic process. The basics can be written in 7 steps:
1. Fig.12.14(a) below shows a thermodynamic process from point A to point B
• The red arrows indicate that, volume increases from VA to VB
♦ So work is done by the gas
• We know that, this work is equal to the area below the red curve
♦ It is marked with magenta hatch lines
 |
| Fig.12.14 |
2. Fig.12.14(b) shows another thermodynamic process between the same points A and B
• But this time the process moves from point B to point A
• The green arrows indicate that, volume decreases from VB to VA
♦ So work is done on the gas
• We know that, this work is equal to the area below the green curve
♦ It is marked with grey hatch lines
3. Let us compare figs (a) and (b). The comparison can be written in 3 steps:
(i) The magenta area is larger than the grey area. That means:
♦ the work done by the gas
♦ is greater than
♦ the work done on the gas
(ii) This is an advantage because, we are receiving a 'net work'
• Obviously,
♦ this net work
♦ will be equal to
♦ the difference between the magenta and grey areas
(iii) The difference between the areas can be obtained by superimposing the grey area onto the magenta area. This is shown in fig.c
• We see that, the difference is equal to the area enclosed by the two curves
• So we can write:
♦ The net work done by the gas
♦ is equal to
♦ The area enclosed between the two curves
4. In fig.c, we see that:
• The process starts at A, moves to B and then moves back to A
♦ The movement from A to B is along the red curve
♦ The movement from B, back to A is along the green curve
5. Change in internal energy:
♦ When the gas is at A, it's internal energy is UA
♦ When the gas is back at A, it's internal energy is again UA
• So the change in internal energy = ΔU = (UA - UA) = 0
6. Now we apply the first law of thermodynamics for the whole process A-B-A:
• We have: ΔU = Q - W
■ Since ΔU is zero, we get:
The net work done by the gas is equal to the net heat absorbed by the gas
7. A cyclic process is the one which can be described using the above steps (4), (5) and (6)
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