In the previous section we saw state variables. In this section we will see internal energy.
• Internal energy can be explained in 4 steps:
1. Consider the system of molecules inside the container that we saw in the previous section. Assume that, they are gaseous molecules. It is shown again in fig.6.5(a) below:
 |
| Fig.6.5 |
♦ 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 do 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 container.
■ This sum is equal to the internal energy of the gas
♦ It is represented by the letter U
4. An important note:
♦ The container may be placed in a moving vehicle.
♦ So the container will be moving.
♦ As a result, the container will have some kinetic energy.
• The kinetic energy acquired by the motion of the container is not taken into account in the calculations related to the U of the system.
Effect of work on internal energy
Let us see how work will increase the internal energy of a system. It can be written in 12 steps:
1. In fig.6.5(b) above, some water is taken in a container.
2. Boundary walls are such that, they do not allow heat to pass through.
• Such walls are called adiabatic walls
3. Record the initial temperature TA
4. Do some mechanical work on the water. This can be done by rotating a set of small paddles.
(Some images of paddle stirrer can be seen here)
• The amount of work done must be 1 kJ
5. When this work is done, measure the final temperature TB
• We will find that, TB is greater than TA
6. We know that, temperature is a measure of the average kinetic energy of the molecules.
• So an increase in temperature is an indication that, kinetic energy of the molecules have increased.
7. But we have seen just above that, kinetic energy of the molecules contribute towards the internal energy of the system
• That means, an increase in temperature is an indication that, the internal energy U of the system has increased.
• We can write:
Paddle stirrer applied → means work is done → temperature increased → means kinetic energy increased → means internal energy increased.
• In short, when the paddle stirrer is applied, internal energy increased.
8. If UA is the initial internal energy and UB is the final internal energy, UB will be greater than UA
• Since there is no passage of energy through the adiabatic walls, all the work will be utilized for increasing the internal energy from UA to UB
♦ So we can write: UB = UA + W
• Then the change in internal energy will be given by: ΔU = (UB – UA) = W
♦ Where W is the work done on the system.
♦ In our present case, W is 1 kJ
9. Consider the same system in fig.6.5(b)
• Supply energy into the system by using an electric immersion heater.
(Some images of electric immersion heater can be seen here)
• The same 1 kJ energy must be supplied.
10. After supplying this much energy, measure the final temperature TB
• We will find that, TB is same as the value obtained in (5).
11. Since TB is the same, UB will also be the same.
12. Here we see that, UB does not depend on the path.
• We used two different paths to apply the 1 kJ energy.
■ In both paths, UB is the same. So U is a state variable.
Effect of heat on internal energy
We have seen how work increases the internal energy. Next we will see how heat increases U. It can be written in 4 steps:
1. In fig.6.5(c) above, some water is taken inside the container.
• This time the walls are not adiabatic. They allow heat to pass through.
2. Bring this container into contact with a hot thermal reservoir.
• A 'thermal reservoir' can be explained in 6 steps:
(i) A thermal reservoir is a body having a large mass m.
(ii) It also has a large specific heat capacity c.
(iii) We know that, the heat absorbed by a body is equal to mcΔT (Details here).
(iv) Since m and c are large for a thermal reservoir, it can absorb a large quantity of heat.
(v) Once the reservoir reaches a high temperature, it will stay at that temperature for a long time.
• This is because, a large quantity of heat will have to be released to bring about even a small drop in temperature.
• That means, a hot thermal reservoir can supply heat with out itself becoming colder.
(vi) Similarly, if a thermal reservoir is cold, it will stay cold for a long time.
• This is because, a large quantity of heat will have to be supplied to bring about even a small rise in temperature.
• That means, a cold thermal reservoir can absorb heat with out itself becoming hot.
3. When we place our system in contact with a hot reservoir, heat will begin to flow into the system.
• Let TA be the initial temperature and TB be the final temperature.
♦ Obviously, TB will be larger than TA
♦ That means, energy of the molecules has increased.
♦ That means, energy of the system has increased.
4. Let Q be the heat entering the system.
• All that heat will be utilized to increase the internal energy to UB
• So we can write: UB = UA + Q
• Then the change in internal energy will be given by: ΔU = (UB – UA) = Q
Combined effect of work and heat
• We have seen the effects of work and heat on the internal energy.
• We have considered work and heat separately.
• Now we will see their combined effect. It can be written in 9 steps:
1. In many practical situations, work and heat will be acting simultaneously on the system.
• In such situations, we can write: UB = UA + Q + W
2. That means, both the ‘supplied heat’ and 'work done' are utilized for increasing the internal energy from UA to UB
3. Then the change in internal energy is given by: ΔU = (UB – UA) = Q + W
• Thus we can write Eq.6.1: ΔU = Q + W
4. Eq.6.1 is the mathematical statement of the first law of thermodynamics
■ The law states that: Energy of an isolated system is constant.
5. Let us see how this law can be explained. It can be written in 4 steps:
(i) Recall that, in an isolated system, there is no flow of energy into or out of the system
• So Q and W will be zero.
(ii) Then Eq.6.1 will become: ΔU = (0 + 0) = 0
(iii) That means, there is no change in internal energy.
(iv) ‘No change in internal energy’ means that: internal energy is a constant
• This indicates that, energy cannot be created.
6. Another explanation can also be given. It can be written in 2 steps:
(i) We saw that, Q and W, if supplied, will be utilized to increase the internal energy
(ii) None of Q or W goes underutilized.
• This indicates that, energy cannot be destroyed
7. Combining (5) and (6), we can write:
■ Energy can neither be created nor destroyed
8. It is difficult to find the exact value of U of a system because, we will need to know the energies of each molecule in the system.
• However, the exact U is not required for solving problems. What we need is the difference ΔU.
• The first law of thermodynamic gives us an easy method to find this ΔU
9. In the Eq.6.1, we must follow IUPAC sign conventions:
• Sign convention for W:
♦ W is taken as positive if work is done on the system.
✰ This is because, work done on the system will cause an increase in U.
♦ W is taken as negative if work is done by the system.
✰ This is because, work done by the system will cause a decrease in U.
• Sign convention for Q:
♦ Q is taken as positive if heat is added to the system.
✰ This is because, heat added will cause an increase in U.
♦ Q is taken as negative if heat is released by the system.
✰ This is because, heat released will cause a decrease in U.
Let us see a solved example
Solved example 6.1
Express the change in internal energy of a system when
(i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ?
(ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?
(iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be ?
Solution:
Part (i):
• Heat is zero.
• Work, W is done on the system.
♦ This W will cause an increase the internal energy, U. So Uf = Ui + W
• Thus we get: ΔU = [Uf - Ui] = [(Ui + W) - Ui] = W
• Only an adiabatic wall can prevent the passage of heat, while allowing work to be done.
Part (ii):
• Work is zero.
• Heat, Q is taken out from the system.
♦ So this Q will cause a decrease in U. So Uf = Ui - Q
• Thus we get: ΔU = [Uf - Ui] = [(Ui - Q) - Ui] = -Q
• The wall is of thermally conducting type.
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 Uf = Ui - W + Q
• Thus we get: ΔU = [Uf - Ui] = [(Ui - W + Q) - Ui] = Q - W
• This is a closed system.
• We have also seen the first law of thermodynamics
• The above two topics are discussed in detail in Physics classes also.
♦ Those notes can be seen in section 12 and section 12.1 of the physics notes.
• The reader is advised to obtain a good understanding about those notes also.
• Consider a system consisting of some gas molecules.
♦ They are kept inside a cylinder-piston arrangement.
• If we supply heat, the gas will expand and do work on the piston.
• The quantity of work done by the gas can be mathematically calculated.
♦ The method is explained in section 12.2 of the physics notes.
• The method will have to be slightly modified depending upon the manner in which the thermodynamic process is carried out.
• There are mainly four types of thermodynamic processes. They are:
Isothermal process, Isobaric process, Isochoric process, Adiabatic process
• The details of these processes and work done in each of them are explained in section 12.3 of the physics notes.
Thus we get the following results:
1. Work done in an isothermal process = $\mathbf\small{\rm{n R T\,\,ln \frac{V_B}{V_A}}}$
2. Work done in an isobaric process = P(VB-VA)
3. Work done in an isochoric process = 0
4. Work done in an adiabatic process = $\mathbf\small{\rm{\frac{1}{1\, - \,\gamma}\left[P_B V_B \, - \, P_A V_A \right]}}$
Let us see some solved examples:
Solved example 6.2
Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into
a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Solution:
1. Given that:
PA = 10 atm, VA = 2 litres, VB = 10 litres
2. Given that the process is isothermal
• In an isothermal process, there is no change in temperature
• So TB = TA = 25 C
3. If there is no change in temperature, there will be no change in internal energy U
4. Applying the first law of thermodynamics, we get:
UB = UA + Q - W
• W is given a negative sign because, when gas expands, work is done by the gas
5. Given that: the gas expands into vacuum
• That means, there is no force opposing the expansion of the gas
• That means, the gas does not have to exert any force against surroundings
• If no force is exerted, no work is done
• We can write: work done by the gas is zero
6. So the result in (4) becomes:
UB = UA + Q - 0
UB - UA = Q
7. But from (3), we have: UB - UA = 0
• So the result in (6) becomes: 0 = Q
8. So we can write:
In this problem, both Q and W are zero
Solved example 6.3
Consider the same expansion in solved example 6.2, but this time against a constant external pressure of 1 atm. Find Q and W
Solution:
1. The gas has to do work against a constant pressure of 1 atm
• 1 atm = 1 × 1.013 × 105 N m-2
2. Since the pressure is constant, it is an isobaric process
• The work is given by: W = P(VB-VA)
• Substituting the known values, we get:
W = 1 × 1.013 × 10 5 × (10 - 2) × 10-3 = 810.4 joules
Solved example 6.4
Consider the expansion given in solved example 6.2, for 1 mol of an ideal gas conducted reversibly.
Solution:
1. For an isothermal process, the work is given by:
$\mathbf\small{\rm{W=n R T\,\,ln \frac{V_B}{V_A}}}$
2. Given that: n = 1, T = (273+25) = 298 K
• Substituting the known values, we get:
$\mathbf\small{\rm{W=1 \times 8.314 \times 298 \times \,\,ln \frac{10}{2}}}$ = 3987.49 joules
Solved example 6.5
One
mol of a mono atomic ideal gas expands isothermally from state A to
state B as shown in the fig.6.6 below. Calculate the work if the
temperature is 298 K
 |
| Fig.6.6 |
Solution:
1. Given that:
n = 1, PA = 2 atm, PB = 1 atm, VA = 22.7 L, T = 298 K
2. For an ideal gas, we have: PAVA = PBVB
• Substituting the known values, we get: VB = 45.4 L
3. For isothermal process, we have : $\mathbf\small{\rm{W=n R T\,\,ln \frac{V_B}{V_A}}}$
• Substituting the known values, we get:
$\mathbf\small{\rm{W=1 \times 8.314 \times 298 \times \,\,ln \frac{45.4}{22.7}}}$ = 1717.32 joules
4. In this problem, the gas expands. So W is negative
Thus we get: W = -1717.32 J
In the next section, we will see enthalpy
PREVIOUS
CONTENTS
NEXT
Copyright©2021 Higher Secondary Chemistry. blogspot.in - All Rights Reserved
No comments:
Post a Comment