Content Strategy Manager
Thermodynamic processes are the paths we can take to bring a thermodynamic system from its initial state to its final state. Thermodynamic systems can be understood as an enclosed space in the universe whose state can be indicated in terms of parameters like volume, temperature, pressure, internal energy, etc. There are 3 types of thermodynamic systems: open, closed, and isolated. The state of these systems can be changed through different processes. The different thermodynamic processes include:
- Isobaric process
- Isochoric process
- Isothermal process
- Adiabatic process
- Quasi-static process
| Table of Content |
Key Terms: Thermodynamics, Thermodynamic Process, Isochoric, Adiabatic, Isothermal, Isobaric, Thermodynamic System
What are Thermodynamic Processes?
[Click Here for Sample Questions]
A thermodynamic process involves a change from one type of equilibrium microstate to another type of system. The process can be interpreted by the initial and final states of the system.
- Temperature pressure, energy, and volume of the system have to be considered as parameters for the initial state. After the completion of the time for which we are observing the system, we can measure the same parameters to figure out the final state of the system.
- Usually, these changes are governed by energy transfer which leads to work done on the system or by the system.
- An example of a thermodynamics process is increasing the pressure of gas in a container where the temperature is constant.
Types of Thermodynamic Processes
[Click Here for Previous Year Questions]
Many different parameters can be used to express the state of any thermodynamic system such as temperature, pressure and volume, and internal energy. The value of any of these parameters can be found if two out of three parameters are fixed. We can do so by using the equation, PV = RT. The state of the system does not remain constant and can be changed through different thermodynamics processes such as:
-
Isothermal Process
The Isothermal Process is the process in which the temperature remains constant in the system. Some heat transfer does take place but it is typically at an extremely slow rate which enables it to attain thermal equilibrium.
Since, W = ∫PdV
From Gas Law,
PV = nRT
P = nRT/V
Putting the value of P, we get:
W = nRT \(\int\limits_{VA}^{VB} {dV \over V}\)
W = nRT ln \({VB \over VA}\)
If VB > VA, the work done will be positive.
If VB < VA, the work done is negative.
Internal energy is constant in such systems because the temperature is constant, ΔU = 0. Therefore, according to the first law of thermodynamics,
Q = ΔU + W
Therefore, Q = W.
Read More: MCQ on Thermodynamics
-
Isobaric Process
A thermodynamics process in which the pressure of the system does not change with time is called an isobaric process.
Work done, W = P (VB - VA)
If ΔV = positive, the work done will be positive.
If ΔV = negative, the work done will be negative.
-
Isochoric Process
In isochoric thermodynamic process, the volume of the system does not change. An example of such a process is gas inside a closed container. The work done by such a system is zero. However, if the system is heated, there will be changes in the internal energy of the system.
According to the first law of thermodynamics,
ΔQ = ΔU + W,
Where Q = heat transferred
and ΔU = change in internal energy.
Therefore, ΔQ = ΔU.
-
Adiabatic Process
In an adiabatic process, no heat is exchanged between the surroundings and the system.
PVγ = K or constant
Since, W = ∫PdV
Using the value of P,
W = \(K\int\limits_{Vi}^{Vf} {dV \over V^\gamma}\)
W = \(K {{Vf^{1-γ} - Vi^{1-γ}} \over {1-γ}}\)
According to the first law of thermodynamics,
Q = ΔU + W
Since, Q= 0, ΔU = -W
Check Out: Thermodynamics Important Questions
All thermodynamic processes can be basically categorised into:
-
Reversible Process
The thermodynamics process that can be reversed, that is, brought back into the initial state by very small changes in the properties of the system is a reversible process. During a reversible process, there is no increase in the entropy of the system and the whole system is in perfect equilibrium with the surroundings.
The work done by the system in a reversible process,
W = ∫PdV,
where P = pressure of the system and V = volume.
On heating, ice cubes can be converted into the water while when we freeze water we get ice cubes. Hence, this is a reversible process.
-
Irreversible Process
During a thermodynamics process, if there is an increase in the entropy of the system then the system cannot return to its original state. The surroundings will also go through some thermodynamic changes and will not be able to return to their original or initial state and this type of process is called irreversible process.
-
Cyclic Process
A cyclic process is a thermodynamics process in which a system goes through a cyclic change during which all its properties change periodically but return to their initial state ultimately and the total change in the internal energy of the system remains zero.
Also Read:
Thermodynamic System
[Click Here for Sample Questions]
The three types of thermodynamic systems are:
- Open system: In an open system, mass and energy are free to flow in and out of the system.
- Closed system: In a closed system, only energy can flow in and out while mass is restricted within the system.
- Isolated system: An isolated system, as the name suggests, is a system in which both mass and energy do not interact with the outside world. Therefore, there is no flow of mass or energy in and out of this system.
In such systems, multiple thermodynamics processes can take place. These processes are classified into various types.
Things to Remember
- Thermodynamic processes are the paths we can take to bring a thermodynamic system from its initial state to its final state.
- Reversible Process: The thermodynamics process can be reversed or brought back into the initial state
- Irreversible Process: If there is an increase in the entropy of the system then the system cannot return to its original state.
- Cyclic Process: A system goes through a cyclic change during which all its properties change periodically but return to their initial state ultimately.
- Isothermal Process: The process in which the temperature remains constant in the system.
- Isobaric Process: The process in which the pressure of the system does not change with time.
- Isochoric Process: The process in which the volume of the system does not change.
- Adiabatic Process: The process in which no heat is exchanged between the surroundings and the system.
Sample Questions
Ques. If a gas system increases ideal gas with a volume of 10 moles from 1 to 20 Litres at a temperature of 0°C. Calculate the amount of work that is done by the system. (2 marks)
Ans. Since this is an isothermal process,
Work done W = nRTlnV2/V1
W = 10 X 8.314 X 273 X ln 20/1
= 10 X 8.314 X 273 X 3
= 6.79 X 104 J
Ques. What are reversible processes? State an example of a reversible process. (2 marks)
Ans. The process where the system is brought back into the initial state by very small changes in its properties is known as a reversible process. Spring Extension is an example of a reversible process.
Ques. The volume of 1m of gas is doubled at atmospheric pressure. Calculate what will the work be the work done at constant pressure? (2 marks)
Ans. Work done in isobaric process, W= P(Vf-Vi) = 1X 105(2-1) = 1 X 105 J
Ques. What is bulk modulus for an isobaric process? (2 marks)
Ans. Since bulk modulus is a ratio of the increase in the pressure to the relative decrease in the volume, the bulk modulus in an isobaric process is zero as the pressure remains unchanged.
Ques. The thermal energy of a gas decreases when it expands in a vessel. What is the process involved here? (2 marks)
Ans. Since the thermal energy of a gas decreases when it expands in a vessel, the process involved is an adiabatic process.
dU + dW = dQ
dU = - dW
Ques. 500gm volume of water is heated from a temperature of 30°C to 60°C. Calculate water’s internal energy change, ignoring the slight expansion of water? (specific heat of water 4184 J/kg.K) (3 marks)
Ans. Since the change in the volume of water is only very small we can consider the process as isochoric. In the isochoric process, the work done by a system is zero. Therefore the only change is going to be in the internal energy because of the heat transfer.
ΔU = Q = msv ΔT
Mass of water = 500 g or 0.5 kg
The change in temperature = 30K
The heat Q = 0.5×4184×30 = 62.76
Ques. The PV diagram for a thermodynamic system is given in the figure below. In the cyclic process that is shown below, calculate the total work done. (3 marks)
Ans. The work done here will be negative as the closed curve is anticlockwise. The work done on the gas can be calculated by the area under the curve BC. The total work done by the system will be calculated by the area under DA.
Area under the curve BC = Area of rectangle BC12 = 1 × 4= − 4J
Area under the curve DA = 1×2= +2J
Hence, the net work done in cyclic process = −4 + 2= −2 J
Ques. 500 g volume of water is heated from a temperature of 30°C to 60°C. Find out the change in water’s internal energy if we ignore the slight water expansion? (Water’s specific heat: 4184 Jkg-1K-1) (3 marks)
Ans. When 500 g water is heated from a temperature of 30°C to 60°C, there is just a negligible amount of change in the volume. Hence, this process can be regarded as isochoric.
So, the work done by the system: 0
ΔU = Q = ms.ΔT
m = 500 g = 0.5 kg
ΔT = 30 K
Therefore, Q = 0.5*4.184*30
= 62.76 KJ
Ques. An ideal gas that is initially at 30 K goes through an isobaric expansion at a pressure of 2.50 kPa. The volume of the gas then increases from 1m3 to 3m3. 12.5 kJ heat is transferred to the gas, then what is the final temperature? (2 marks)
Ans. For an ideal gas
P1V1/T1= P2V2/T2
For an isobaric process, P1 = P2
Hence, T2 = 3T1
= 900 K
Ques. A 0.5 mole of gas expands isothermally at temperature 300 K from 2L to 6L. Calculate:
i) Work done by the gas.
ii) Heat added to the gas.
iii) Final pressure of the gas.
(Value of gas constant, R = 8.31J mol-1K-1) (5 marks)
Ans.
- Work done in isothermal expansion, W = nRT ln (VB/VA)
W = 0.5 mol X 8.31J/mol.K X 300 K In (6L/2L)
W = 1.369 kJ
- According to the first law of thermodynamics, the work is done by the heat transferred.
Therefore, heat added to the gas, Q = W = 1.369kJ
- In a isothermal process,
PiVi+ Pf Vf = μRT
⇒Pf = μRT/Vf
⇒Pf = 0.5mol X 8.31J/mol.K X 300K/ 6X10-3m3
⇒Pf = 207.75 kPa
Therefore, the final pressure of the gas is 207.75 kPa
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:
Comments