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Test: Basic Concepts of Thermodynamics & Applications (June 10) - JEE MCQ


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Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 1

110 joule of heat is added to a gaseous system, whose internal energy is 40 J. Then the amount of external work done is 

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 1

Concept:

  • The first law of thermodynamics is a restatement of the law of conservation of energy. It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.
  • When heat energy is supplied to a thermodynamic system or any machine. 
  • Two things may occur:
    • The internal energy of the System or machine may change.
    • The system may do some external work.

According to the first law of Thermodynamics:

ΔQ = ΔW + ΔU

Where ΔQ = Heat supplied to the system or heat exchange, ΔW = work done by the system, ΔU = change in internal energy of the system 

Explanation:

Given that, ΔQ = 110 J, ΔU = 40 J

According to the  first law of thermodynamics:

ΔQ = ΔW + ΔU

ΔU = ΔQ - ΔW

40 J = 110 J - ΔW

ΔW = 110 - 40 = 70 J

Then the amount of external work done is 70 J.

Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 2

The heat given to an ideal gas in isothermal conditions is used to:

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 2

CONCEPT:

First Law of Thermodynamics:

  • It is a statement of conservation of energy in the thermodynamical process.
  • According to it heat given to a system (ΔQ) is equal to the sum of the increase in its internal energy (ΔU) and the work done (ΔW) by the system against the surroundings.

i.e ΔQ = ΔU + ΔW          [∴ ΔW = p ΔV]

  • It makes no distinction between work and heat as according to it the internal energy (and hence temperature) of a system may be increased either by adding heat to it or doing work on it or both.

EXPLANATION:

  • When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the change is known as an isothermal process.
  • As we know that, the internal energy of the system is a function of temperature alone, so in the isothermal process, the change in internal energy is zero.

⇒ ΔQ = 0 + ΔW =  ΔW

  • Therefore, the heat given to an ideal gas in isothermal conditions is used to do external work. Hence option 2 is correct.
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Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 3

In thermodynamics ___________ is not a state variable.

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 3

CONCEPT:

  • State variables are defined as the thermodynamical variables which depend only on the initial and final state of a thermodynamical system
    • These variables don't depend on how the thermodynamical system changed itself from the initial to the final state.
    • Temperature, Pressure, Internal energy, and Density are the examples of state variables.
    • State variables are also known as state functions.
  • Path variables are defined as the thermodynamical variables which depend on the way in which the thermodynamical system achieved the initial and final states.
    • Heat, Work is examples of Path variables

EXPLANATION:

  • Internal energypressuredensity, and enthalpy are examples of state variables. Since they depend only on the final and initial states of the thermodynamical system.
  • Heat is a measure of the amount of energy present in a thermodynamical system. As the amount of energy changes the heat present in the system changes. Hence heat is the path variable. Therefore option 4 is correct answer.
Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 4

In which Thermodynamic process is there no flow of heat between the system and the surroundings?

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 4

In an adiabatic process, there is no flow of heat between the system and the surroundings. This means that the system is thermally isolated, preventing any exchange of heat energy. Instead, any changes in the system's internal energy occur solely due to work done on or by the system.

Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 5

If the temperature of the source is increased, the efficiency of the Carnot engine

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 5

CONCEPT:

  • Carnot engine: The theoretical engine which works on the Carnot cycle is called a Carnot engine.
    • It gives the maximum possible efficiency among all types of heat engines.
    • The part of the Carnot engine which provides heat to the engine is called a heat source.
    • The temperature of the source is maximum among all the parts.
    • The part of the Carnot engine in which an extra amount of heat is rejected by the engine is called as a heat sink.
    • The amount of work which is done by the engine is called as work done.

The efficiency (η) of a Carnot engine is given by:

Where TC is the temperature of the sink, TH is the temperature of the source, W is work done by the engine, Qin is the heat given to the engine/heat input and QRis heat rejected.

EXPLANATION:

The efficiency (η) of the Carnot engine is given by:

η = 1 - TC/TH

  • Here if TH increases, the value of TC/TH decreases, and hence the value of (1 - TC/TH) increases.
  • If the temperature of the source (TH) is increased then the efficiency of the Carnot engine increases. So option 1 is correct.
Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 6

Melting Point of pure ice is-

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 6

Explanation:

  • The melting point is the temperature at which a solid turns to a liquid
  • The melting point at which ice turns into liquid is 0Oc or 273K.
  • Regelation is the phenomenon of ice melting under pressure and refreezing when the pressure is reduced.

Thus, the melting point of pure ice is 273K.

Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 7

Which of the following is an intensive variable?

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 7

CONCEPT:

  • Intensive Property: These are the properties of the system which are independent of mass under consideration. 
    • For example: Pressure, Temperature, density, composition
  • Extensive Properties: The properties which depend on the mass of the system under consideration. 
    • For example: Internal Energy, Enthalpy, Mass, Volume, Entropy, weight

EXPLANATION:

  • From above it is clear that, the temperature is an intensive variable. Therefore option 2 is correct.

NOTE:

  • All specific properties are intensive properties. For e.g. specific volume, specific entropy, etc
Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 8

The system in which there is no exchange of matter, energy or work with the environment is called _________.

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 8

Key Points

Isolated system:

  • In an Isolated system, there is no mass and energy interaction across the system boundary i.e. Interaction between the system and the surroundings is absent. Therefore, the mass and the energy of the isolated system is fixed e.g. Universe.
  • When a body of material starts from a non-equilibrium state of inhomogeneity or chemical non-equilibrium and is then isolated, it spontaneously evolves towards its own internal state of thermodynamic equilibrium.
  • It is not necessary that all aspects of internal thermodynamic equilibrium be reached simultaneously; some can be established before others. For example, in many cases of such evolution, internal mechanical equilibrium is established much more rapidly than the other aspects of the eventual thermodynamic equilibrium.
Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 9

Which variable is held constant in Charles's Law?

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 9

CONCEPT:

Charles law:

  • If the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

i.e. V ∝ T
or V/T = constant

EXPLANATION:

  • From above it is clear that in Charles's Lawpressure remains constant. Therefore option 4 is correct.
Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 10

What is the source temperature of the Carnot engine in K required to get 70% efficiency?

Given sink temperature = 27 °C

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 10) - Question 10

CONCEPT:

  • Carnot engine: A theoretical thermodynamic cycle proposed by Leonard Carnot. It gives the estimate of the maximum possible efficiency that a heat engine during the conversion process of heat into work and conversely, working between two reservoirs, can possess.
    • So practically and theoretically there can not be any engine with more efficiency than Carnot engine.

The efficiency of the Carnot's Heat engine is given by:

where Tis the temperature of the cold reservoir/sink and Th is the temperature of the hot reservoir/source.

  • The efficiency of this type of engine is independent of the nature of the working substance and is only dependent on the temperature of the hot and cold reservoirs.

CALCULATION:

Given that sink temperature T= 27°C = 300K

η = 70% = 0.7


Th = 300/0.3 = 1000 K 

So the correct answer is option 1.

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