Thermodynamics is the study of heat, work, temperature, and their relationship to energy, entropy, and the physical properties of matter.
Key Concepts:
Thermal Equilibrium: Two systems are in thermal equilibrium when they have the same temperature.
Zeroth Law of Thermodynamics:In simpler terms, if object A is in thermal equilibrium with object C, and object B is also in thermal equilibrium with object C.
First Law of Thermodynamics:
This law states that energy is conserved. The heat added to a system equals the change in its internal energy plus the work done by the system.
Second Law of Thermodynamics:This law governs the direction of natural processes.It says that the disorder or randomness of an isolated system always increases.
Thermodynamic Processes
Isothermal Process: A process occurring at constant temperature.
Adiabatic Process:A process occurring without heat exchange with the surroundings.
Isobaric Process:A process occurring at constant pressure.
Isochoric Process: A process occurring at constant volume.
Heat Engines and Refrigerators:
Heat Engine: Converts heat energy into mechanical work.
Refrigerator:Extracts heat from a cold reservoir and rejects it to a hot reservoir. Its performance is measured by the coefficient of performance (COP), which is the ratio of heat extracted to work input.
1. A geyser heats water flowing at the rate of 3.0 litres 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 ?
Ans : Given:
Volume of water heated per minute = 3.0 liters
Initial temperature of water (Ti) = 27°C
Final temperature of water (Tf) = 77°C
Heat of combustion of fuel = 4.0 × 10^4 J/g
Calculation:
Mass of water heated per minute:
1 liter of water weighs 1 kg.
So, 3 liters/minute = 3 kg/minute = 3000 g/minute
Heat energy required per minute to heat the water:
Heat energy (Q) = mass × specific heat capacity × temperature change
(c) = Specific heat capacity of water
= 4.2 J/g°C
Temperature change (ΔT) = Tf – Ti = 77°C – 27°C = 50°C
Q = 3000 g/min × 4.2 J/g°C × 50°C = 63 × 10^4 J/min
Rate of fuel consumption:
Heat energy produced by fuel = mass of fuel × heat of combustion
63 × 10^4 J/min = mass of fuel × 4.0 × 10^4 J/g
Mass of fuel = (63 × 10^4 J/min) / (4.0 × 10^4 J/g)
= 15.75 g/min
hence, the rate of consumption of the fuel is 15.75 g/min
2. 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; R = 8.3 J mol–1 K–1.)
Ans : *Understanding the Problem:**
We’re asked to calculate the heat required to raise the temperature of a given mass of nitrogen gas at constant pressure.
Solution:
1. Calculate the Number of Moles:
Number of moles (n) = Mass / Molar mass
= (2.0 × 10⁻² kg) / (28 × 10⁻³ kg/mol) ≈ 0.714 moles
2. Use the Specific Heat Capacity at Constant Pressure:
For a diatomic gas like nitrogen, the specific heat capacity at constant pressure (Cp) is approximately 7/2 R.
Cp = (7/2)R = (7/2) × 8.3 J/mol K ≈ 29.05 J/mol K
3. Calculate the Heat Energy:
The heat energy (Q) required to raise the temperature (ΔT) of a substance is given by:
Q = nCpΔT
Substituting the values:
Q = 0.714 mol × 29.05 J/mol K × 45 K
≈ 934 J
Therefore, approximately 934 Joules of heat must be supplied to the nitrogen gas.
3. Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Ans : a) Thermal Equilibrium and Temperature Distribution:
While it might seem intuitive that two bodies at different temperatures would reach a mean temperature when brought into thermal contact, this is not always the case. The final equilibrium temperature depends on the specific heat capacities and masses of the two bodies. If one body has a significantly larger heat capacity or mass, it will exert a greater influence on the final temperature, pulling it closer to its initial temperature.
(b) Coolant with High Specific Heat:
A coolant with a high specific heat capacity can absorb a large amount of heat without a significant increase in its own temperature. This property is crucial for preventing overheating in chemical or nuclear plants. By absorbing heat efficiently, the coolant can maintain a relatively stable temperature, ensuring optimal operation and preventing damage to the plant’s components.
(c) Air Pressure in a Car Tyre:
As a car moves, the air molecules within the tire gain kinetic energy due to the increased motion. This increased kinetic energy results in an increase in the average speed of the air molecules, which in turn leads to an increase in the pressure exerted by the air on the tire’s inner walls.
(d) Climate of a Harbour Town vs. a Desert Town:
Harbour towns tend to have more moderate climates compared to inland desert towns at the same latitude. This is due to the influence of the ocean. Water has a high specific heat capacity, which means it can absorb and release large amounts of heat with relatively small temperature changes. This helps to regulate the temperature of coastal areas, making them less prone to extreme temperature fluctuations. In contrast, desert regions lack the moderating influence of large bodies of water, leading to more extreme temperature variations.
4. 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 ?
Ans : **Understanding the Problem:**
We have a cylinder of hydrogen gas that is compressed to half its original volume, while keeping the system insulated.
Determine factor by which the pressure increases.
Solution:
1. Identify the Process:
Since the system is insulated, no heat is exchanged with the surroundings. This is an **adiabatic process**.
2. Use the Adiabatic Process Equation:
we use following formula for an ideal gas undergoing an adiabatic process
P₁V₁^γ = P₂V₂^γ
Where:
* P₁ and V₁ =
initial pressure and volume, respectively
* P₂ and V₂ =
final pressure and volume, respectively
* γ = adiabatic index
(for a diatomic gas like hydrogen, γ ≈ 1.4)
3. Find the Pressure Factor:
We’re given that the final volume (V₂) is half the initial volume (V₁):
V₂ = 0.5V₁
Substituting this into the adiabatic process equation:
P₁V₁^1.4 = P₂(0.5V₁)^1.4
Simplifying:
P₂ / P₁ = (1/0.5)^1.4
≈ 2.64
Therefore, the pressure of the gas increases by a factor of approximately 2.64.
5. 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)
Ans : Understanding the Problem:
We are given two scenarios involving changes in internal energy (ΔU), heat (ΔQ), and work (ΔW).
In both cases, we need to find the missing variable using the first law of thermodynamics.
First Scenario (Adiabatic Process):
ΔQ = 0 (no heat exchange)
ΔW = -22.3 J (work is done on the system)
Using the first law of thermodynamics:
ΔQ = ΔU + ΔW
0 = ΔU – 22.3 J
Therefore, ΔU = 22.3 J.
Second Scenario:
ΔQ = 9.35 cal = 39.3 J (heat is added to the system)
ΔU = 22.3 J (from the first scenario)
ΔW = ?
Using the first law of thermodynamics again:
ΔQ = ΔU + ΔW
39.3 J = 22.3 J + ΔW
Therefore, ΔW
= 17.0 J.
Conclusion:
In the first scenario, The system gains 22.3 J of internal energy due to the work done on it.
In the second scenario, the work done by the system is 17.0 J.
6. 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 ?
Ans : Understanding the Problem:**
We have two cylinders connected by a stopcock. Initially, one cylinder (A) is filled with gas at STP, while the other (B) is evacuated. When the stopcock is opened, the gas expands to fill both cylinders. We need to analyze the changes in pressure, internal energy, and temperature.
Solution:
(a) Final Pressure:
Since the final volume is doubled (both cylinders combined), and the number of moles of gas remains constant, the final pressure will be halved according to Boyle’s law.
Therefore, the final pressure in both cylinders will be:
P_final = P_initial / 2 = 1 atm / 2
= 0.5 atm
(b)Change in Internal Energy
The system is thermally insulated, meaning no heat is exchanged with the surroundings. Therefore, the process is adiabatic. For an ideal gas undergoing an adiabatic process, the internal energy change (ΔU) is given by:
ΔU = (nR / (γ – 1)) * (T₂ – T₁)
Where:
* n denote number of moles of gas
* R = gas constant
* γ = adiabatic index
for an ideal gas,
γ = Cp
———
Cv
* T₁ and T₂ =
initial & final temperatures, respectively
Since the internal energy of an ideal gas depends only on its temperature, and the temperature remains constant in an isothermal expansion (as we’ll see in part (c)), the change in internal energy is zero:
ΔU = 0
(c) Change in Temperature:The process is adiabatic, and the gas expands into a vacuum. This is an isothermal expansion, as there is no work done on or by the gas. Therefore, the temperature remains constant.
ΔT = 0
(d) Intermediate States and P-V-T Surface:
The intermediate states of the system, as the gas expands to fill both cylinders, will not lie on the P-V-T surface of an ideal gas. This is because the process is not quasi-static, meaning the system is not in thermodynamic equilibrium during the expansion. The pressure and temperature may vary rapidly and non-uniformly within the system as the gas expands.
7. An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Ans : *Understanding the Problem:**
We have a system that is receiving heat energy at a certain rate and doing work at a certain rate.
Determine rate of change of internal energy of the system.
Solution:
1.
Apply the First Law of Thermodynamics:
ΔU = Q – W
2.
Calculate the Rate of Change of Internal Energy:
We are given the rate of heat supply (dQ/dt) and the rate of work done (dW/dt). We need to find the rate of change of internal energy (dU/dt):
dU/dt = dQ/dt – dW/dt
Substituting the given values:
dU/dt = 100 W – 75 W
= 25 W
Therefore, the internal energy of the system is increasing at a rate of 25 Watts.
8. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13)
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
Ans : Understanding the Problem:
We are given a pressure-volume (PV) diagram showing a process from point D to point F.
Determine work done by the gas during this process.
Calculation:
Change in Pressure (ΔP):
ΔP = EF = 5.0 atm – 2.0 atm = 3.0 atm = 3.0 × 10^5 Nm^-2
Change in Volume (ΔV):
ΔV = DF = 600 cc – 300 cc = 300 cc = 300 × 10^-6 m^3
Work Done by the Gas:
The work done by the gas during a process is represented by the area under the PV curve.
In this case, the area of the trapezoid DFEF represents the work done.
we use formula
Area of a trapezoid
= (1/2) × (sum of parallel sides) × (perpendicular distance between them)
Applying this formula to the trapezoid DFEF:
Work Done (W) = (1/2) × (DF + EF) × ΔP
Substituting the values:
W = (1/2) × (300 × 10^-6 m^3 + 0) × (3.0 × 10^5 Nm^-2)
= 45 J
Therefore, the work done by the gas from point D to point F is 45 Joules.
