In the previous section, we saw the second law of thermodynamics applied to heat engine and refrigerator. In this section, we will see some solved examples related to the topics that we saw in this chapter
Solved example 12.11
A geyser heats water flowing at the rate of 3.0 liters per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ?
Solution:
1. Volume of water heated in one minute = 3 liter
2. So mass of water heated in one minute = [3(l) × 1000 (g l-1)] = 3000 g = 3.0 kg
3. Heat absorbed in one minute =
Mass in one minute × Specific heat capacity × Change in temperature
4. Specific heat capacity of water = 4180 J kg-1 K-1
5. Change in temperature = (77 - 27) = 50 °C
6. Substituting the known values in (3),we get:
Heat absorbed in one minute =
3.0 (kg) × 4180 (J kg-1 K-1 ) × 50 K = 627000 J
7. Given that, one gram of the fuel gives 4.0 × 104 J
So number of grams of fuel required in one minute
= 627000 (J) ÷ 4.0 × 104 (J g-1) = 15.7 g
Solved example 12.12
What amount of heat must be supplied to 2.0 × 10-2 kg of nitrogen (at room
temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular
mass of N2 = 28 g; R = 8.3 J mol-1 K-1)
Solution:
Method 1:
1. We have: Heat required =
Mass × Specific heat capacity × Change in temperature
2. Mass of nitrogen = 2.0 × 10-2 kg
3. From data book, we get:
Specific heat capacity of nitrogen (at constant pressure, cp) = 1040 J kg-1 K-1
4. Change in temperature = 45 °C
5. Substituting the known values in (1),we get:
Heat required =
2.0 × 10-2 (kg) × 1040 (J kg-1 K-1 ) × 45 (K) = 936 J
Method 2:
1. We have: Heat required =
Number of moles × Molar heat capacity × Change in temperature
2. Mass of nitrogen = 2.0 × 10-2 kg
• Molar mass of N2 = 28 gram
♦ So number of moles of N2 in 2.0 × 10-2 kg
= [2.0 × 10-2 (kg) ÷ 28 × 10-3 (kg)] = 0.714
3. For a diatomic gas, we have:
Molar heat capacity (at constant pressure, cp) = 7⁄2 R (We will derive this relation in the next chapter)
• Substituting for R, we get: cp = [7⁄2 × 8.3 (J mol-1 K-1)] =
4. Change in temperature = 45 °C
5. Substituting the known values in (1),we get:
Heat required =
0.714 (mol) × [7⁄2 × 8.3 (J mol-1 K-1)] × 45 (K) = 933.3 J
Solved example 12.13
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?
Solution:
1. From the description it is clear that, it is an adiabatic process
• We have the relation for adiabatic process: $\mathbf\small{\rm{P_1 V_1^\gamma = P_1 V_2^\gamma}}$
• For a diatomic gas, $\mathbf\small{\rm{\gamma}}$ = 1.4
2. Given that initially, the gas is at STP
♦ So P1 = 1 atm
• One mole of any ideal gas will occupy 22.4 L at STP
♦ So V1 = (3 × 22.4) L
3. Given that the gas is compressed to half it's original volume
♦ So V2 = 0.5V1
4. Substituting the known values in (1), we get:
$\mathbf\small{\rm{1 \; (atm) \times (3 \times 22.4 \; (L))^{1.4} = P_2 \times (0.5 \times 3 \times 22.4 \; (L))^{1.4}}}$
⇒ P2 = 2.64 × P1
Solved example 12.14
In changing the state of a gas adiabatically from an equilibrium state A to another
equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)
Solution:
• There are two processes in this problem
♦ The first process is an adiabatic process
♦ The second process is not adiabatic
1. Let A and B be the initial and final states of the gas. Let W1 be the work done
• For any thermodynamic process, we have: UB = UA + Q - W
• For the first process,
Q = 0 and W1 is positive (since work is done on the gas)
• Thus we get: UB = (UA + W1) = (UA + 22.3 J)
⇒ UB - UA = 22.3 J
2. For the second process, we have:
UB = UA + (9.35 × 4.19) - W2
(Here W2 is negative because, work is done by the gas)
⇒ UB - UA = (9.35 × 4.19 J) - W2
3. What ever be the process, UA and UB will be the same
• So (UA - UB) will also be the same
• Equating the results in (1) and (2), we get:
UB - UA = 22.3 J = (9.35 × 4.19 J) - W2
⇒ W2 = [(9.35 × 4.19 J) - 22.3] = 16.9 J
Solved example 12.15
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
(a) What is the final pressure of the gas in A and B ?
(b) What is the change in internal energy of the gas ?
(c) What is the change in the temperature of the gas ?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?
Solution:
• For an ideal gas, we have: $\mathbf\small{\rm{\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}}}$
• For our present problem, it is given that the entire system is thermally insulated
♦ So no heat will enter or leave the system
Part (c): Cylinder B is initially a vacuum
• The gas in A, expands and fills into B. For this, no work is necessary because, the gas is expanding against vacuum
• Since no work is done on/by the gas, the temperature remains the same
♦ That is., T1 = T2
♦ So change in temperature = (T2 - T1) = 0
Part (a):
• Since T1 = T2, we get: P1V1 = P2V2
• Given that:
♦ P1 = 1 atm and V1 = V2
♦ So final volume available = 2V1
• So we get: 1 × V1 = P2 × 2V1
⇒ P2 = 0.5 atm
Part (b):
• Since work done is zero and heat exchange is also zero, there will be no change in internal energy
• That is., change in internal energy = 0
Part (d):
• This is a rapid process
• The intermediate P and V values will be fluctuating
• So those values will not lie on the P-V-T surface
(see fig.12.4 in the first section of this chapter)
Solved example 12.16
A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Part (a):
• Efficiency is given by: $\mathbf\small{\rm{\eta = \frac{Output}{Input}}}$
• Substituting the values, we get:
$\mathbf\small{\rm{\eta = \frac{5.4 \times 10^8 (J)}{3.6 \times 10^9 (J)}}}$ = 0.15 = 15%
Part (b):
Energy wasted = [Input energy - Output work]
= [3.6 × 109 - 5.4×108] = 30.6 ×108 J
Solved example 12.17
An electric heater supplies heat to a system at a rate of 100 W. If system performs
work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:
1. 100 W is 100 J s-1
• So energy absorbed by the system in one second, Q = 100 J
2. Work done by the system in one second = 75 J
3. We can write: UB = UA + 100 J - 75 J
• Work is given a negative sign because, it is done by the system
4. Thus we get:
Change in internal energy in one second
= (UB - UA) = (100 - 75) = 25 J
Solved example 12.18
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.25). Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F
 |
| Fig.12.25 |
Solution:
1. In the initial process, work is done by the gas while expanding from VD to VE
• This work will be equal to:
The area between segment DE and the x-axis
2. In the second process, work is done on the gas while compressing it from VE to VF
• This work is equal to:
The area between segment EF and x-axis
3. So the net work done by the gas is equal to:
The area of the triangle DEF
♦ Base of the triangle = (5 - 2) = 3
♦ Altitude of the triangle = (600 - 300) = 300
♦ So area of the triangle = 1⁄2 × 3 (m3) × 300 (N m-2) = 450 J
In the next
chapter, we will see kinetic theory
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