Thermodynamics Important Questions

Very Short Answer Questions (1 Mark Questions)

Ques. On what grounds is thermodynamics based? 

Ans. The foundations of thermodynamics are the law of conservation of energy and the fact that heat transfers from a hot body to a cold body.

Ques. Is it possible to liquefy a gas at any temperature just by increasing the pressure? 

Ans. No. When the temperature of the gas is below its critical point, only then it can it be liquefied by pressure alone.

Ques. What does the term "universal gas constant" mean? 

Ans. The universal gas constant denotes the work done per mole per Kelvin by (or on) a gas.

Ques. What exactly do you mean when you say thermodynamical system?

Ans. An assemblage of a high number of particles is referred to as a thermodynamical system (atoms or molecules).

Ques. Explain what thermodynamical variables mean. 

Ans. Thermodynamical variables are the parameter that describes the system’s equilibrium state.

Ques. What does "positive" and "negative" work mean to you? 

Ans. Positive work done by the system is referred to as such, whilst negative work is referred to as such.

Ques. Why is it that a body with high reflectivity is a poor emitter?

Ans. This is due to the fact that a body with high reflectivity is a poor heat absorber, and poor absorbers are poor emitters.

Ques. Define heat pump?

Ans. A heat pump is a device that removes heat using mechanical work.

Ques. Is it possible for two isothermal curves to cross one other?

Ans. No, it is not possible to intersect two isothermal curves because if they will intersect there will be two temperature values.

Read More: Thermodynamic Processes

Short Answer Questions (2 Marks Questions)

Ques. What is the thermodynamics zeroth law?

Ans. If the thermodynamic systems are both in thermal equilibrium with a third thermodynamic system C, then the systems are also in thermal equilibrium, according to the Zeroth law of Thermodynamics.

Ques. Why is heat energy lower than visible light energy?

Ans. An electromagnetic device's energy is calculated as follows:

E = hf

f = wave frequency; h = Planck's constant Thermal radiation has a lower frequency than visible light, hence its energy is lower.

Ques. When the sky is cloudy on a winter night, it feels warmer than when it is clear. Why? 

Ans. We know that during the day, the earth absorbs heat and then radiates it at night. When the sky is cloudy, the heat radiated by the earth is reflected back, making the earth warmer. When the sky is clear, however, the heat emitted by the planet escapes into space.

Ques. When the piston in a cycle pump is rapidly lowered, the pump heats up. Why? 

Ans. This is due to the fact that some of the work done in pumping gets transformed into heat. When the piston is swiftly lowered, there is no time for heat to transfer from the piston to the surrounding area, causing the pump to overheat.

Ques. Define the isochoric process? What kind of work will be done throughout the procedure? 

Ans. The isochoric process refers to a thermodynamical process in which the volume of the system remains constant. Because the volume of the gaseous system does not change, the work done in this operation will be 0.

Read More: Isobaric Process

Ques. Why is a material's latent heat of vaporization greater than that of fusion?

Ans. When a liquid turns into a gas, the volume expands dramatically, and a great deal of work is required against the surrounding atmosphere. The heat associated with the transition from solid to gas is known as latent heat of vaporization, and thus the answer.

Ques. When constructing thermometers, why is mercury used?

Ans. Mercury is used to make thermometers because it has a large and useful temperature range as well as a consistent rate of expansion.

Long Answer Questions (3 Marks Questions)

Ques. The statements of the Second Law of Thermodynamics by Kelvin and Clausius are identical. Explain? 

Ans. Assume we have an engine that produces a continuous source of work when cooled below the ambient temperature.

This is a breach of Kelvin's promise. Now, if the engine's work is utilized to power a dynamo that generates current, and this current generates heat in a coil immersed in hot water, we've created a machine that allows heat to transfer from a cold body to a hot body without the use of an external agent. This is a contravention of Clausius's assertion. As a result, both statements are equal.

Ques. No practical engine can have a higher efficiency than a Carnot engine operating at the same temperatures. Why? 

Ans. From the following perspectives, a Carnot engine is ideal:

Between the cylinder walls and the piston, there is no friction. The working substance is an ideal gas, which means that the gas molecules have no molecular attraction and are little points.

These requirements, however, cannot be met in a practical engine, and so no heat engine operating between the same two temperatures can have a higher efficiency than a Carnot engine.

Ques. The process of adiabatically altering the state of gas from one equilibrium state A to another equilibrium state B on the system, 22.3 J worth of work is completed. How much net work is done by the system in the latter scenario 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? (Take 1 caI = 4.19 J as an example.)

Ans. Given that dQ = O and dW = -22.3 J in the first adiabatic process (-ve sign indicates here that work is done on the system)

dQ = dU + dW
∴ 0 = dU – 22.3 or dU = 22.3 J, according to the first law of thermodynamics.

In the second phase,

dQ = 9.35 cal = 9.35 × 4.19 J = 39.18 J

dQ = dU + dW

39.18 = 22.3 + dW

Or, dW = 39.18 – 22.3 = 16.88 J

= 16.9 J

According to the first rule of thermodynamics.

First Law of Thermodynamics

Ques. Why is C_p for a gas bigger than C_v?

Ans. When heat is applied to gas while maintaining its volume constant, the entire amount of heat is used to enhance the kinetic energy and thus the temperature of the gas. However, if the pressure is to be maintained, more heat is required to raise the temperature of the same amount of gas by the same amount as the gas must work against pressure during an expansion (volume is now increased). C_p > C_v, as defined by the definitions of C_p and C_v.

Ques. Calculate the energy in Joule that must be given to freeze one kg of water, assuming the residential refrigerator is a reversible engine operating between the melting point of ice and the room temperature of 27°C. L = 80 cal g^-1 when the water temperature is 0°C.

Ans. T₁ = 27 + 273 = 300 K in this case.

T₂ = 0 + 273 = 273 K

m = 1 kg = 1000 g

L = 80 cal g^-1

Q₂ = mL = 1000

80 cal = 8 × 10⁴ cal to be eliminated heat

Using the relation

Q₁ / Q₂ = T₁ / T₂, we get

Q₁ = T₁ / T₂ × Q₂ = 300 / 273 × 8 × 10⁴

= 87912.1 cal

Energy required to be supplied,

W = Q₁ – Q₂ = (87912.1 – 80000) cal

= 7912.1 cal

= 7912.1 × 4.2 J = 33230.8 J

Read More: Difference between Isothermal and Adiabatic Process

Very Long Answer Questions (5 Marks Questions)

Ques. Explain whether the procedures listed below are reversible.

1. Waterfall
2. Electrolysis

Ans. (a) The process of water falling is not reversible. The majority of the potential energy of the waterfall is turned into kinetic energy of the water, and a portion of it is transferred into heat and sound as it strikes the earth. It is impossible to convert the heat and sound produced by water's K.E. into potential energy, allowing the water to return to its original height.

(b) Electrolysis is a process that involves the use of electricity. If the electrolyte provides no resistance to the current passage, the process is reversible. When the direction of the current is reversed, the direction of ion motion is reversed as well.

Ques. Explain the importance of the second law of thermodynamics.

Ans. The first law of thermodynamics establishes an equivalence between heat and mechanical energy, but it remains silent on the direction of energy change.

Heat travels from a body at a higher temperature to a body at a lower temperature, but not from a body at a lower temperature to a body at a higher temperature, as we all know. Why? Why is a heat engine's efficiency less than one? When a chum is used to swirl the water in a beaker, it heats up. However, no mechanical effort is obtained when a chum is immersed in hot water in a beaker. What is the reason for this? The second law of thermodynamics can be used to answer these questions.

The major statements of the second law of thermodynamics are as follows:

(i) Clausius statement: It is impossible for a self-acting machine to transfer heat from a body at a lower temperature to a body at a higher temperature without the assistance of an external source. It is clear from this statement that heat can never flow from a body at a lower temperature to a body at a higher temperature unless external work is performed on it.

This assertion is consistent with observations in other disciplines of physics, such as the fact that electric current never flows from a lower potential conductor to a higher potential conductor unless external effort is performed on it. Similarly, a body cannot move from a lower to a higher surface unless it is worked on by an external source.

(ii) Kelvin statement: It is impossible to provide a constant source of work by cooling a body to a temperature lower than the coldest part of the environment. If the temperatures of the source and sink are the same in a Carnot engine, the engine will not run because if it does, work will be done and the source will cool down below the ambient temperature.

(iii) Kelvin-Plank statement: It is impossible to build a machine that converts all of the heat obtained from a source into work completely in a cycle. This remark implies that the working substance in a heat engine can never transform all of the heat absorbed from the source into work.

It’s critical to direct some of the heat away from the sink. For the conversion of heat into work, both the source and the sink must be present. Because not all of the heat is converted into work, the engine’s efficiency is always less than one.

Read More: Planck Equation: Planck's Constant & Black Body Radiation

Ques. State the first law of thermodynamics and prove C_p – C_v = R using this law.

Or

Prove Mayer's C_p – C_v = R relationship.

Or

Determine the link between a gas's specific heat under constant pressure and volume.

Ans. The first law of thermodynamics states: If a system can perform external work, the total quantity of heat energy delivered to it is equal to the sum of the energy spent doing work (by the system) plus the increase in internal energy of the system. C_p and C_v Relationship: Give one mole of an ideal gas an amount of heat Q at a constant volume so that its temperature rises by T.

Q = 1. C_v.T, (Q = mST) Q = C_v.T...

Because the volume of the gas remains constant, the gas does no external work.

∴ ΔW = 0

From first law of thermodynamics, we have
ΔQ = ΔW + ΔU
or
ΔQ = 0 + ΔU
or
ΔQ = ΔU …(2)
From eqns. (1) and (2), we get
ΔU = C_v.ΔT …(3)

Allow one mole of an ideal gas to receive an amount of heat ΔQ under constant pressure, causing its temperature to rise by ΔT. 

ΔQ = 1.C_p.ΔT …(4)

If ΔV be the increase in the volume of the gas at constant pressure P, then the work done by the gas

ΔW = P.ΔT,

[∵ Work = Force × Distance = Force Area × Volume = Pressure × Volume]

Now, from the first law of thermodynamics,

ΔQ = ΔW + ΔU

Or

ΔQ = PΔV + ΔU …(5)

From eqns. (4) and (5), we get

PΔV + ΔU = C_pΔT

Now, put the value of ΔU from eqn. (3),

We get PΔV + C_vΔT = C_pΔT

Or

(C_p – C_v)ΔT = PΔV …(6)

The ideal gas equation is given by

PV = RT

Or

PΔV = RΔT, (∵ P is constant)

Putting the value of PΔV Fin eqn. (6),we get

(C_p – C_v)ΔT = RΔT

Or

C_p – C_v = R

Which is the required relation.

Ques. Shikhar explained to his pal that heat cannot be transferred from a colder to a warmer body. His friend Raman was taken aback because the first law of thermodynamics contained no such explanations for the direction in which change can occur. They finally went to their physics teacher to find out why.

1. What are the values they display?
2. What explanation did the teacher give?

Ans. (i) Curiosity, group discussion, and a desire to understand the scientific rationale are the values they demonstrate.

(ii) The first law of thermodynamics, as described by the teacher, has a constraint. At higher temperatures, heat is transferred from the body to lower temperatures. The second law of thermodynamics states that ds 0 means that the system cannot advance in the direction of decreasing entropy/probability. As a result, heat cannot travel from a colder body to a hotter body on its own.