Chapter 12.1 - The First Law of Thermodynamics

In the previous section we saw quasi static process and reversible process. In this section, we will see the first law of thermodynamics.

• First we will see internal energy U. It can be explained in 4 steps:
1. Consider the gas inside the cylinder that we saw in the previous section. It is shown again in fig.12.5 below:

Fig.12.5

• The molecules of the gas will be always in motion.
    ♦ Due to the motion, molecules will be having kinetic energies.
• The molecules will be colliding with each other.
    ♦ So their velocities will always be changing.
    ♦ So kinetic energies will also be changing.
2. But for our present discussion, such changes in kinetic energies does not matter.
• What is important is this:
    ♦ Each molecule will be having some amount of kinetic energy.
• In addition to kinetic energies, the molecules will be having:
    ♦ Potential Energy.
    ♦ Energy due to vibratory motions.
    ♦ Energy due to rotational motions.
3. We take the sum of all the above four types of energies of all the molecules in the cylinder.
■ This sum is equal to the internal energy of the gas
    ♦ It is represented using the letter U
4. An important note:
    ♦ The 'cylinder containing the gas' may be placed in a moving vehicle.
    ♦ So the cylinder will be moving.
    ♦ As a result, the cylinder will have some kinetic energy.
• The kinetic energy acquired by the motion of the cylinder is not taken into account in the calculations related to the U of the gas.

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Now we can learn about the first law of thermodynamics. It can be written in 15 steps:
1. We have seen the basic details about U. There are various methods to calculate the exact value of U. We will see them in higher classes.
■ In scientific and engineering problems, we are more interested in calculating ΔU, which is the change in internal energy.
2. Suppose that:
    ♦ U_i is the initial internal energy of the gas.
    ♦ U_f is the final internal energy of the gas.
• Then we have: ΔU = U_f – U_i
■ This ΔU can be determined easily by applying the first law of thermodynamics
    ♦ The following steps from (3) to (7) explains how this is done:
3. Suppose that, the gas in the cylinder has an initial internal energy of U_i joules.
• Let the gas be heated. Let Q joules of energy flow into the gas.
4. When the gas is heated, it will expand and push the piston upwards.
• Since the piston is moved upwards, we can say:
    ♦ Gas is doing work on the piston.
• Let the gas do a work of W joules.
5. Now, when the gas acquires Q joules of heat, it’s total internal energy becomes (U_i + Q)
6. But when the gas does work, it loses some energy.
    ♦ In our present case, W joules will be lost.
7. So the total internal energy becomes U_i + Q – W
• So we can write: Final internal energy, U_f = U_i + Q – W
• So ΔU = [U_f – U_i] = [(U_i + Q – W) – U_i] = Q – W
• Thus we get an equation:
Eq.12.1: ΔU = Q – W
• This is the mathematical form of the first law of thermodynamics.
• The law states that:
The change in internal energy of a system ΔU is the difference between heat supplied to the gas and the work done by the gas.
(In our present case, 'all the molecules of the gas in the cylinder' is the ‘system’)
8. The flow of heat Q into the system can be achieved by any one of the four ways:
    ♦ Place a burner below the cylinder.
    ♦ Place the cylinder in contact with a hot body.
    ♦ Immerse the cylinder in boiling water.
    ♦ Ignite the gas (if the gas is combustible) using an electric circuit.
9. Flow of heat Q out of the system can also be achieved easily:
    ♦ Immerse the cylinder in a box of ice.
10. We have to pay special attention to the signs of each of the three items: ΔU, Q and W
The following steps from (11) to (15) explain this:
11. Suppose that Q₁ joules of heat flowed into the gas
• But after some time Q₂ joules of heat flowed out of the gas.
• Then the net heat is the difference between Q₁ and Q₂
• That is., ΔQ = Q₁ – Q₂
12. Q₁ being larger or smaller:
• If Q₁ is larger, we can say: The gas has a net heat gain.
    ♦ Because, more heat flowed in and less heat flowed out.
    ♦ Mathematically, [ΔQ = (Q₁ – Q₂)] will be positive.
    ♦ So we can use a positive value of Q in Eq.12.1
• If Q₁ is smaller, we can say: The gas has a net heat loss.
    ♦ Because, less heat flowed in and more heat flowed out.
    ♦ Mathematically, [ΔQ = (Q₁ – Q₂)] will be negative.
    ♦ Then we must use the negative value of Q in Eq.12.1
13. Similar steps can be written for work also.
• Suppose that W₁ joules of work was done by the gas.
• But after some time W₂ joules of work was done on the gas.
• Then the net work is the difference between W₁ and W₂
• That is., ΔW = W₁ – W₂
14. W₁ being larger or smaller:
• If W₁ is larger, we can say: The gas does a net work.
    ♦ Because, more work is done by the gas and less work was done on the gas.
    ♦ Mathematically, [ΔW = (W₁ – W₂)] will be positive.
    ♦ So we can use a positive value of W in Eq.12.1
    ♦ But in the equation, a -ve sign is already present before W.
    ♦ So the positive work will be deducted:
          ✰ ΔU = Q - (+W) = Q - W
• If W₁ is smaller, we can say: The gas has a net work done on it.
    ♦ Because, less work is done by the gas and more work is done on the gas.
    ♦ Mathematically, [ΔW = (W₁ – W₂)] will be negative.
    ♦ Then we must use the negative value of W in Eq.12.1
    ♦ But in the equation, a -ve sign is already present before W.
   ♦ So the positive work will be added:
          ✰ ΔU = Q - (-W) = Q + W
• This is obvious because, a ‘net work done on the gas’ will surely increase the internal energy of the gas.
15. We must have a clear understanding about the above sign convention.
• However, while doing problems, it is better to ‘consider the effects’ rather than just following the sign convention.
• We can ‘consider the effects’ as follows:
    ♦ A net gain of heat will increase the internal energy of the gas.
    ♦ A net loss of heat will decrease the internal energy of the gas.
    ♦ A net work done by the gas will decrease the internal energy of the gas.
    ♦ A net work done on the gas will increase the internal energy of the gas.

The following two solved examples demonstrate the facts written in (15):
Solved example 12.1
Express the change in internal energy of the system when
(i) No heat is absorbed by the system from the surroundings, but work of W joules is done on the system
(ii) No work is done on the system, but Q joules of heat is taken out from the system and given to the surroundings
(iii)  W joules of work is done by the system and Q joules of heat is supplied to the system
Solution:
Part (i):
• Heat is zero.
• Work, W is done on the system.
    ♦ This W will cause an increase the internal energy, U. So U_f = U_i + W
• Thus we get: ΔU = [U_f - U_i] = [(U_i + W) - U_i] = W
Part (ii):
• Work is zero.
• Heat, Q is taken out from the system.
    ♦ So this Q will cause a decrease in U. So U_f = U_i - Q
• Thus we get: ΔU = [U_f - U_i] = [(U_i - Q) - U_i] = -Q
Part (iii):
• Work, W is done by the system.
    ♦ This W will cause a decrease in U.
• Heat Q is supplied to the system.
    ♦ This Q will cause an increase in U.
• So U_f = U_i - W + Q
• Thus we get: ΔU = [U_f - U_i] = [(U_i - W + Q) - U_i] = Q - W

Solved example 12.2
In a thermodynamic process, 40 J of heat is supplied to the system. When this heat is supplied, the system does 10 J of work. After some time, 4 J of work is done on the system. When this work is done, 25 J of heat is released from the system into the surroundings. What is the change in internal energy of the system after the whole process?
Solution:
Stage (1):
• Heat of 40 J is supplied to the system.
    ♦ This heat will cause an increase in U.
• Work of 10 J is done by the system.
    ♦ This W will cause a decrease in U.
• So U_f at the end of stage 1 = U_f(1) = (U_i + 40 - 10) = U_i + 30
    ♦ This U_f(1) is the U_i for stage 2.
Stage 2:
• Work of 4 J is done on the system.
    ♦ This W will cause an increase in U.
• Heat of 25 J is released from the system.
    ♦ This Q will cause a decrease in U.
• So U_f at the end of stage 2 = U_f(2) = (U_f(1) + 4 - 25) = (U_f(1) - 21)
= (U_i + 30 - 21) = U_i + 9
Net change in U at the end of the two stages:
ΔU = (U_f(2) - U_i) = (U_i + 9 - U_i) = 9 J

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Next, we will see the relation between the following three items:
Internal internal energy U, temperature T and heat Q. It can be written in 3 steps:
1. Experiments show that, the temperature T is directly proportional to the internal energy U of the gas. That is., T ∝ U
• So, if U_i is the initial internal energy and T_i is the initial temperature, we can write: T_i ∝ U_i
• Similarly, if U_f is the final internal energy and T_f is the final temperature, we can write: T_f ∝ U_f
2. The temperature of the gas can change from T_i to T_f,
    ♦ even if zero Q is added to the system.
    ♦ even if zero Q is taken away from the system.
• We can prove this in 4 steps:
(i) Let a work of W joules be done on the gas.
(ii) Let Q be equal to zero. That is., no heat is supplied or taken away.
(iii) Then we have: U_f = U_i + W
So clearly, U_f is greater than U_i
(iv) From (1), we have: T_i ∝ U_i and T_f ∝ U_f
So, since U_f is greater than U_i, T_f will be greater than T_i
3. So we see that, it is not necessary for a heat-flow (either inwards or outwards) to change the temperature.

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In the previous section, we have seen the details about P-V Diagram.
Let us see how the three quantities (ΔU, Q and W) are related to the P-V diagram. It can be written in 5 steps:
1. Fig.12.6(a) below shows a P-V Diagram.
    ♦ It is the diagram related to the gas in fig.12.5 above.
• The initial state of the gas is marked as A(V_A, P_A)
• The final state B(V_B, P_B) can be any where on the diagram.
    ♦ Position of B will depend on the heat content Q and work content W.

Fig.12.6

2. Draw a vertical dashed line through A.
    ♦ The portion to the right of the vertical dashed line is shaded with red color.
    ♦ The portion to the left of the vertical dashed line is shaded with green color.
3. Sign of work W:
• This can be determined in 2 steps:
(i) If B is any where in the red region, V_B will be greater than V_A
    ♦ That means, the volume has increased.
    ♦ That means, gas has done work on the piston.
    ♦ Let this work be W.
    ♦ So considering the sign, it will be: – W
(ii) If B is any where in the green region, V_B will be less than V_A
    ♦ That means, the volume has decreased.
    ♦ That means, work has been done on the gas.
    ♦ Let this work be W.
    ♦ So considering the sign, it will be: + W
4. Sign of ΔU:
This can be determined in 9 steps:
(i) Let the gas in cylinder in fig.12.5 be an ideal gas.
• Consider the ideal gas equation: PV = nRT
• If there is no leakage of gas in fig.12.5, n will be a constant.
• We know that R is always a constant.
• So we can write: PV ∝ T
(ii) We have seen that, the temperature T is proportional to the internal energy U.
• That means:
    ♦ An increase in T is an indication that U has increased.
    ♦ A decrease in T is an indication that U has decreased.
(iii) But from (i),
    ♦ An increase in T is an indication that PV has increased.
    ♦ A decrease in T is an indication that PV has decreased.
(iv) So it is clear that:
    ♦ An increase in PV is an indication that U has increased.
    ♦ A decrease in PV is an indication that U has decreased.
(v) Draw vertical and horizontal dashed lines through A.
• Those lines will divide the graph into four portions.
    ♦ The top right portion is shaded with magenta color.
    ♦ The bottom left portion is shaded with cyan color.
    ♦ This is shown in fig.12.6(b) above.
(vi) If B is in the magenta portion,
    ♦ P_B will be greater than P_A
    ♦ V_B will be greater than V_A
• Then,
    ♦ P_BV_B will be greater than P_AV_A
    ♦ That means, PV value has increased.
• If there is an increase in PV, we know that U_f will be greater than U_i
    ♦ So [ΔU = (U_f – U_i)] will be positive.
(vii) If B is in the cyan portion,
    ♦ P_B will be less than P_A
    ♦ V_B will be less than V_A
• Then,
    ♦ P_BV_B will be less than P_AV_A
    ♦ That means, PV value has decreased.
• If there is a decrease in PV, we know that U_f will be less than U_i
    ♦ So [ΔU = (U_f – U_i)] will be negative.
(viii) If B is in the black portion at top left,
    ♦ P_B will be greater than P_A
    ♦ V_B will be less than V_A
• Then we cannot say which one of (P_AV_A) and (P_BV_B) is larger.
In this situation, we will need to pick the actual P_A, V_A, P_B  and V_B values from the graph.
• Here an interesting situation can arise:
    ♦ Suppose that, P_B = 2P_A and V_B = 0.5V_A
    ♦ Then P_BV_B = (2P_A × 0.5V_A) = P_AV_A
    ♦ That means, PV value at B is same as that at A.
    ♦ In such a situation, Uf will be same as Ui.
    ♦ So ΔU will be zero.
(ix) If B is in the black portion at bottom right,
    ♦ P_B will be less than P_A
    ♦ V_B will be greater than V_A
• Then we cannot say which one of (P_AV_A) and (P_BV_B) is larger.
In this situation also, we will need to pick the actual P_A, V_A, P_B  and V_B values from the graph.
5. So from the above four steps, we can determine the sign of W and ΔU.
• Once we know the sign of those two items, we can find the sign of Q
• The following solved example will demonstrate the procedure.

Solved example 12.3
In a thermodynamic process, B is in the cyan region in fig.12.6(b) above. What will be the signs of ΔU, W and Q.
Solution:
1. B is in the cyan region.
    ♦ So B is at the left side of A.
    ♦ So V_B is less than V_A
• That means, work is done on the gas.
    ♦ So it is a positive value.
2. B is in the cyan region.
    ♦ So PV has decreased.
• So U_f is less than U_i
    ♦ So [ΔU = (U_f – U_i)] will be negative.
3. We have: U_f – U_i = Q + W
    ♦ From (2), we have: The left side is -ve.
    ♦ From (1), W is positive.
• So we get: (-) = Q + (+)
⇒ (-) + (-) = Q
• So the sign of Q is negative.
• That means, Q has flowed out of the gas.

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In the next section, we will see the method to calculate the work done during a thermodynamic process.

 

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