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Thermodynamic properties are defined as a set of characteristics that explain the state of a system based on its physical characteristics. The thermodynamic properties of a material are classified into the following categories
- Measured properties: These are the properties that are directly accessible from measurements in the laboratory. Examples of measured properties are volume, temperature, and pressure.
- Fundamental properties: These properties of the material are directly related to the fundamental law of thermodynamics. Examples of fundamental properties are internal energy and enthalpy.
- Derived properties: These properties are the specific combinations of derived and measured properties. Examples of derived properties are Gibbs energy, enthalpy, etc.
Thermodynamics is a branch of physics that deals with the study of heat and its transformation to mechanical energy.
- A collection of a large number of molecules confined within certain boundaries such that they have a certain value of pressure, volume, and temperature is called a thermodynamic system.
- Anything outside the thermodynamic system is called surrounding.
Very Short Answers Questions [1 Mark Questions]
Ques. Which among the following are derived properties?
- Temperature
- Gibbs energy
- Volume
- Internal energy
Ans. The correct answer is b. Gibbs free energy
Explanation: Gibbs energy is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant.
Ques. Entropy is
- Fundamental properties
- Measured properties
- Derived properties
- None of the above
Ans. The correct answer is a. Fundamental properties
Explanation: Since entropy is directly related to the fundamental law of thermodynamics, hence it is a fundamental property.
Ques. In thermodynamics, heat and work are
- Extensive thermodynamic variables
- Point functions
- Path functions
- Intensive thermodynamic state variables
Ans. The correct answer is c. Path functions
Explanation: The thermodynamic quantities whose values depend on the initial and final state are known as Path functions.
Ques. What is the formula to find the enthalpy?
- H = E + PV
- H = E – PV
- H = E x PV
- H = E / PV
Ans. The correct answer is a. H = E + PV
Explanation: In a thermodynamic system, enthalpy is the measurement of energy. The formula of enthalpy is given by
H = E + PV
Where
- E is the internal energy
- H is the enthalpy
- P is pressure
- V is volume
Ques. Thermos flask is a
- Isolated system
- Diathermic system
- Only closed system
- Only adiabatic system
Ans. The correct answer is a. Isolated system
Explanation: A thermos flask is an isolated system used to keep things either hot or cold.
Short Answers Questions [2 Marks Questions]
Ques. What are the four categories of thermodynamics?
Ans. The four categories of thermodynamics are
- Statistical Thermodynamics
- Classical Thermodynamics
- Equilibrium Thermodynamics
- Chemical Thermodynamics
Ques. What is meant by the term fundamental thermodynamic properties?
Ans. The properties of the system that are directly related to the fundamental laws of thermodynamics are called fundamental properties.
Examples of fundamental thermodynamic properties are enthalpy and internal energy.
Ques. Define the internal energy of a system. Write its characteristics.
Ans. The internal energy of a system is defined as the sum of the kinetic and potential energies of all the constituent molecules of the system.
The characteristics of internal energy are
- It depends on the state of the system
- It is a state variable that is independent of the path taken to arrive at that state.
- It does not depend on the motion of a system as a whole.
Ques. For a gaseous system, find the change in internal energy if the heat is supplied to the system is 50 J and the work done by the system is 16 J.
Ans. According to the first law of thermodynamics,
ΔQ = ΔU + ΔW
Where
- ΔQ = Heat supplied to or removed from the system
- ΔU = change in internal energy of the system
- ΔW = Work done by or on the system.
Given
- ΔQ = 50 J
- ΔW = 16 J
Therefore, ΔU = ΔQ - ΔW
⇒ ΔU = 50 - 16 = 34 J
Long Answers Questions [3 Marks Questions]
Ques. What are the differences between path functions and point functions?
Ans. The following are the differences between the path functions and point functions
| Path Functions | Point Functions |
|---|---|
| The thermodynamic variables that depend on the path to reach one state from another to define a thermodynamic process are known as the Path function. | The thermodynamic variables which do not depend on the path to reach one state from another to define a thermodynamic process are known as the Point function. |
| The differential of the path functions is inexact | The differential of the point functions is exact. |
| Examples of the path functions are heat and work. | Pressure, temperature, volume, etc. are examples of point functions. |
Ques. What are the differences between entropy and enthalpy?
Ans. The followings are the differences between entropy and enthalpy
| Enthalpy | Entropy |
|---|---|
| It is the measurement of the energy of the system | It is the measurement of the disorders of the system |
| It is given by the sum of internal energy and the product of pressure and volume. | It is the amount of heat transferred reversibly in and out of the system at a given temperature. |
| It is a kind of energy in the system | It is a physical property of the system |
| A thermodynamic system favors minimum enthalpy | A thermodynamic system favors maximum entropy |
Ques. What is the difference between intensive and extensive thermodynamic properties?
Ans. The following are the differences between intensive and extensive thermodynamic properties
| Intensive Property | Extensive Property |
|---|---|
| The properties of a thermodynamic system that are independent of mass are called intensive properties. | The properties of the thermodynamic system that depends on mass are called extensive properties. |
| These properties of thermodynamic systems are not additive in nature. | These properties of thermodynamic systems are additive in nature. |
| Pressure, Temperature, Density, Specific heat capacity, etc. are examples of intensive properties. | Internal energy, enthalpy, entropy, mass, and volume are examples of extensive properties. |
Very Long Answers Questions [5 Marks Questions]
Ques. A girl is running along a beach and she does 4.3 x 105 J of work and give off 3.8 x 105 J of heat.
- What is the change in her internal energy?
- If she does not run but walks, she then gives off 1.2 x 105 J of heat and her internal energy decreases by 2.6 x 105 J. How much work has she done while walking?
Ans. Consider the girl to be a system. She does work on surrounding i.e. Work done by the system on surrounding
ΔW = + 4.3 x 105 J
She releases heat to the surroundings i.e. ΔQ is negative
ΔQ = - 3.8 x 105 J
- Let ΔU be the change in internal energy, then using the first law of thermodynamics,
ΔQ = ΔU + ΔW
⇒ ΔU = ΔQ - ΔW
⇒ ΔU = (-3.8 x 105) - (4.3 x 105)
⇒ ΔU = - 8.1 x 105 J
- While walking, she gives off heat to surrounding
ΔQ = - 1.2 x 105 J
Her internal energy decreases, i.e.
ΔU = - 2.6 x 105 J
Using the first law of thermodynamics,
ΔQ = ΔU + ΔW
⇒ ΔW = ΔQ - ΔU
⇒ ΔW = (- 1.2 x 105) - (- 2.6 x 105)
⇒ ΔW = 1.4 x 105 J
Ques. Define thermodynamic state variables. What are the different types of state variables?
Ans. The equilibrium state of a thermodynamic system can be described completely by some parameters or microscopic variables. These parameters or variables which describe the equilibrium state of the system are called thermodynamic state variables.
Thermodynamic state variables are of two kinds
- Intensive variables: These are the variables that are independent of the size of the system. Examples are pressure, density, and temperature.
- Extensive variables: These are the variables that depend on the size of the system. Examples are volume, mass, internal energy, etc.
Ques. The pressure 1 x 105 N/m2 of the air filled in a vessel is decreased adiabatically so much as to increase its volume three times. Calculate the pressure of air. ℽ for air = 1.4.
Ans. Let the initial pressure of the air filled in the vessel be P and the final pressure be P’
Also, Let V be the initial volume of the air and V’ be the final volume.
Given,
- P = 1 x 105 N/m2
- V’ = 3V
For adiabatic expansion, we have PVℽ = constant
Where ℽ is the ratio of specific heats.
⇒ PVℽ = P’V’ℽ
⇒ 1 x 105 x Vℽ = P’(3V)ℽ
⇒ 105 Vℽ = P’ 3ℽ Vℽ
⇒ P’ = 105 / 3ℽ
⇒ P’ = 105 / 31.4
⇒ log P’ = log 105 - log 31.4
⇒ P’ = 2.148 x 104 N/m2
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