In the previous section we saw the details about isothermal process. In this section, we will see the remaining three processes.
Isobaric process
This can be explained in 8 steps:
1. The word 'iso' in Greek means equal. 'Baros' means pressure.
• So isobaric process is a process in which the pressure remains the same from start to finish.
2. Consider the gas inside the cylinder in fig.12.10(a) below.
• The piston is free to move. But in fig.(a), it is in equilibrium at a certain position.
 |
| Fig.12.10 |
3. Let us assume that the mass of the piston is negligible.
• Then we can write:
♦ The pressure exerted by the gas, P
♦ is equal to
♦ The sum of the 'pressure due to the 5 kg mass' and the 'atmospheric pressure Pa'
4. Let A be the area of cross section of the piston.
• Then the pressure due to the 5 kg mass = 5⁄A
• So we get: P = 5⁄A + Pa
♦ This is how we get the equilibrium mentioned in (2)
5. Now, if we heat the gas slowly (by supplying Q joules of heat), it's pressure will increase and become a new value P1
• But we want the pressure to remain the same at (5⁄A + Pa)
• This can be achieved if the piston is 'free to move'.
• How does the pressure remain constant when the piston is 'free to move' ?
The answer can be written in 3 steps:
(i) When the pressure increases to P1, the gas pushes the piston upwards.
• So volume increases. This is shown in fig.b
(ii) Since more volume is available, the gas molecules have more space to move around.
• They will not hit the walls of the cylinder with the same force.
(iii) That means, the pressure has decreased.
• Thus, even when heat is applied, the pressure remains equal to (5⁄A+Pa)
6. Since the pressure remains the same, the P-V Diagram will be a horizontal line.
This is shown in fig.c
• We know that, work done by the gas will be equal to the area under the P-V diagram.
• In our present case, it is easy to find the area because, it is a rectangular area.
7. In addition to the above graphical method, we can use the ideal gas equation also to calculate the work. It can be explained in 5 steps:
(i) We have: PV = nRT
(ii) Here P, n and R are constants. So we get:
PVA = nRTA and PVB = nRTB
(iii) Subtracting PVA from both sides of 'PVB = nRTB', we get:
PVB - PVA = nRTB - PVA
(iv) But PVA = nRTA
• Using this on the right side of (iii), we get:
PVB-PVA = nRTB - nRTA
⇒ P(VB-VA) = nR(TB-TA)
(v) But P(VB-VA) = P ΔV, is the work done.
• So we get Eq.12.4: Work done in an isobaric process = nR(TB-TA)
8. Let us apply the first law of thermodynamics to an isobaric process. It can be written in 3 steps:
(i) When the gas expands, work is done by the gas.
• Eq.12.1 becomes: ΔU = Q - W
• So we get: ΔU = Q - P(VB-VA)
• The heat Q can be calculated using Eq.11.9 that we saw in the previous chapter (Details here).
Q = nCPΔT
• Thus it is possible to calculate ΔU
(ii) When the gas contracts, work is done on the gas.
• Eq.12.1 becomes: ΔU = Q + W
• So we get: ΔU = Q + P(VA-VB)
• The heat Q can be calculated using Eq.11.9 as before.
(iii) Once we obtain W and Q, we can easily calculate ΔU.
Solved example 12.4
Find the change in internal energy of 1 gram of boiling water when it is changed from liquid state to gaseous state.
Solution:
1. Take 1 gram of 'water at 100 °C' in an open vessel.
• This water is at 100 °C. So it has an initial internal energy, Ui
• If we supply heat to this water, it will vaporize into steam.
2. Let us see the pressure acting on the system.
• Our system is the 1 g of water at 100 °C
• It is in an open vessel. So the atmospheric pressure Pa is compressing it.
• We can assume that, a 'mass less imaginary piston' is applying a pressure of Pa on the water.
3. Since heat is being supplied, the water gradually turns to steam.
• This steam pushes the imaginary piston upwards.
♦ So pressure will not build up below the piston.
♦ The system will be under the constant pressure of Pa all the while.
• So it is an isobaric process.
4. For an isobaric process, we can calculate the work done easily.
• In our present case, the equation will be: Pa(Vf - Vi)
♦ Vi is the initial volume occupied by 1 gram of water.
✰ When the pressure is Pa, this volume is equal to 1 cm3
♦ Vf is the volume occupied by the steam produced from 1 gram water.
✰ When the pressure is Pa, this volume is equal to 1671 cm3
• Thus W = Pa(Vf - Vi) = 1.013 × 105 × (1671 - 1) × 10-6 = 169.2 J
5. Next we want the quantity of heat supplied to the 1 gram water.
• We know that, water at boiling point will require the latent heat to vaporize (Details here).
• This latent heat will be equal to: (Mass of water × Latent heat of water).
= (1 gram × 2256 J/gram) = 2256 J
6. Now we can apply the first law of thermodynamics.
• We have: ΔU = Q - W
• Substituting the values, we get: ΔU = 2256 - 169.2 = 2086.8 J
7. Remember that, we supplied 2256 J heat.
• But a major portion (2086.8 J) of that heat is used to increase the internal energy of the system.
• The system did a work of only 169.2 J
• If instead of water, the system was a gas, we could obtain more work.
Isochoric process
This can be explained in 4 steps:
1. In an isochoric process, volume remains the same.
• This can be achieved by locking the piston at a suitable position.
2. Since there is no change in volume, work, W will be zero.
• This zero work can be confirmed graphically also:
♦ Since the volume is constant, the P-V Diagram will be a vertical line.
♦ A vertical line has zero area below it.
3. When W is zero, we can apply the first law of thermodynamics easily.
• We have: ΔU = Q + W
⇒ ΔU = Q + 0
⇒ ΔU = Q
4. We can find Q easily:
• Since the gas is at a constant volume, the heat absorbed by it will be equal to mCV(TB-TA)
♦ Where m is the mass of the gas.
♦ CV is the specific heat of the gas at constant volume.
♦ TA is the initial temperature and TB is the final temperature.
• If this much heat is supplied to the gas, all that heat will be utilized to increase the internal energy of the gas.
• If this much heat is released from the gas, there will be a corresponding decrease in the internal energy of the gas.
Adiabatic process
This can be explained in 7 steps:
1. In an adiabatic process, there is no flow of heat into or out of the system.
• So we assume that, there is an imaginary insulating layer on these two areas:
♦ the inside surface of the cylinder (green layer in fig.12.11 below)
♦ the underside of the piston (pink layer in fig.12.11 below)
 |
| Fig.12.11 |
2. Since there is no heat flow in either direction, we can apply the first law of thermodynamics easily:
• We have: ΔU = Q + W
⇒ ΔU = 0 + W
⇒ ΔU = W
3. So we can write:
• If we do any work on the gas, all those work will be utilized to increase the internal energy of the gas.
• If the gas does any work, that work will be at the expense of the internal energy.
4. Note that, in an adiabatic process, neither pressure nor volume is a constant.
• So the P-V Diagram will be a curve.
(Recall that, if P is constant, the diagram will be a horizontal line. If V is constant, the diagram will be a vertical line. In our present case, both P and V vary)
• We can use calculus to find the area below the curve
♦ We will need the equation of the curve.
(Recall the steps that we did in the case of isothermal process. See fig.12.9 of the previous section)
5. For an adiabatic process, the equation is: $\mathbf\small{\rm{PV^{\gamma}=A \, Constant}}$
♦ Where $\mathbf\small{\rm{\gamma=\frac{C_P}{C_V}}}$
6. Now we can apply calculus. We get the following result:
The area below the curve = $\mathbf\small{\rm{\frac{1}{1\, - \,\gamma}\left[P_B V_B \, - \, P_A V_A \right]}}$
7. Thus we get Eq.12.5: $\mathbf\small{\rm{W=\frac{1}{1\, - \,\gamma}\left[P_B V_B \, - \, P_A V_A \right]}}$
8. We saw that, an ideal gas subjected to an adiabatic process will satisfy the equation in (5).
• But any ideal gas will satisfy the equation PV = nRT also.
• Based on this, we can derive another equation for W. It can be written in 5 steps:
(i) PAVA = nRTA
(ii) PBVB = nRTB
(iii) Substracting PBVB from both sides of (i), we get:
PAVA – PBVB = nRTA – PBVB
(iv) But from (ii), we have: PBVB = nRTB
• So the right side of (iii) can be changed:
PAVA – PBVB = nRTA – nRTB
⇒ PAVA – PBVB = nR(TA – TB)
(v) So Eq.12.5 can be modified as: $\mathbf\small{\rm{W=\frac{-1}{1\, - \,\gamma}\left[nR(T_A \, - \, T_B) \right]}}$
■ Thus we get Eq.12.6: $\mathbf\small{\rm{W=\frac{nR(T_A \, - \, T_B)}{\gamma\, - \,1}}}$
• Where:
♦ W is the work done in an adiabatic process.
♦ TA and TB are the initial and final temperatures.
9. We can derive another interesting equation related to adiabatic processes. It can be written in 4 steps:
(i) In the previous step (8), we have used subscripts A and B.
• ‘A’ denotes the initial state of the gas and B denotes the final state of the gas.
♦ We successfully calculated the work done from A to B.
• We can use subscripts 1, 2, 3 etc., also.
♦ They will denote the different states of the gas.
(ii) Any ideal gas will satisfy the equation PV = nRT
• n and R are constants. So we get:
P1V1 = nRT1 and P2V2 = nRT2
• Dividing the first equation by P2V2, we get: $\mathbf\small{\rm{\frac{P_1 V_1}{P_2 V_2}=\frac{nRT1}{P_2 V_2}}}$
• But P2V2 = nRT2
So we get: $\mathbf\small{\rm{\frac{P_1 V_1}{P_2 V_2}=\frac{nRT_1}{nRT_2}=\frac{T_1}{T_2}}}$
(iii) When this ideal gas is subjected to an adiabatic process, it will satisfy the relation $\mathbf\small{\rm{PV^{\gamma}=A \, Constant}}$ also
So we get: $\mathbf\small{\rm{P_1 V_1^{\gamma}=P_2 V_2^{\gamma}}}$
$\mathbf\small{\rm{\frac{P_1}{P_2}=\left ( \frac{V_2}{V_1} \right )^{\gamma}}}$
(iv) Substituting this in (ii), we get:
$\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma} \times \frac{V_1}{V_2}=\frac{T_1}{T_2}}}$
■ Thus we get Eq.12.7:
$\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma - 1}=\frac{T_1}{T_2}}}$
Let us see two solved examples
Solved example 12.5
During an adiabatic process, a gas expands from state 1 to state 2. The initial temperature T1 is 292 K. Calculate the final temperature T2, if the volume expansion ratio is 1.28. Given that 𝛄 = 1.40
Solution:
1. We have Eq.12.7: $\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma - 1}=\frac{T_1}{T_2}}}$
2. Given that $\mathbf\small{\rm{\frac{V_2}{V_1}}}$ = 1.28
3. Substituting all the known values in Eq.12.7, we get:
$\mathbf\small{\rm{(1.28)^{1.40 - 1}=\frac{292}{T_2}}}$
• Thus we get: T2 = 265 K
Solved example 12.6
In an automobile engine, the initial values related to the gas are as follows:
T1 = 293 K, P1 = 1.00 × 105 N/m2 and V1 = 240 cm3. This gas is compressed adiabatically to a final volume V2, which is 40 cm3. Find the values of P2 and T2. Also find the work done on the gas. Given that, 𝛄 = 1.40
Solution:
Part (a). To find P2:
• For an adiabatic process, we have: $\mathbf\small{\rm{P_1 V_1^{\gamma}=P_2 V_2^{\gamma}}}$
⇒ $\mathbf\small{\rm{P_2=P_1\left ( \frac{V_1}{V_2} \right )^{\gamma}}}$
• Substituting all the known values, we get:
$\mathbf\small{\rm{P_2=1.00 \times 10^5 \times \left ( \frac{240}{40} \right )^{1.4}}}$ = 12.3 × 105 N/m2
Part (b). To find T2
• We have Eq.12.7: $\mathbf\small{\rm{\left ( \frac{V_2}{V_1} \right )^{\gamma - 1}=\frac{T_1}{T_2}}}$
• Substituting all the known values, we get:
$\mathbf\small{\rm{\left(\frac{40}{240} \right )^{1.40 - 1}=\frac{293}{T_2}}}$
• Thus we get: T2 = 600 K
Part (c). To find W
• We have Eq.12.5: $\mathbf\small{\rm{W=\frac{1}{1\, - \,\gamma}\left[P_2 V_2 \, - \, P_1 V_1 \right]}}$
• Substituting all the known values, we get:
$\mathbf\small{\rm{W=\frac{1}{1\, - \,1.4}\left[(12.3) (40) \, - \, (1)(240) \right]\times 10^5 \times 10^{-6}}}$
• Thus we get: W = -63 J
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