First Law of Thermodynamics Important Questions

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The first law of thermodynamics is a statement of conservation of energy in thermodynamical processes. 

It states that the heat given to a system is equal to the sum of the increase in its internal energy and the work done by the system against the surroundings.

Mathematically, the equation of first law of thermodynamics is given by

ΔQ = ΔU + ΔW

Where

  • ΔQ is the heat energy given to the system
  • ΔU is the change in the internal energy of the system
  • ΔW is the work done by the system.

A few limitations of the first law of thermodynamics are

  • It does not tell us about the direction of the transfer of heat
  • It does not say anything about whether the process is spontaneous or not.
  • The reverse process is not possible.

The first law of thermodynamics

The first law of thermodynamics


Very Short Answers Questions [1 Mark Questions]

Ques. The equation of the first law of thermodynamics is

  1. Internal Energy = Heat added plus work done
  2. Internal Energy = Heat rejected into work done
  3. Internal Energy = Heat added into work done
  4. Internal Energy = Heat added divided by work done

Ans. The correct answer is a. Internal Energy = Heat added plus work done

Explanation: The first law of thermodynamics can be expressed mathematically as:

ΔU = Q + W

Where

  • ΔU is the change in internal energy
  • Q is heat
  • W is work done

Ques. The first law of thermodynamics deals with

  1. Conservation of momentum
  2. Conservation of pressure
  3. Conservation of mass
  4. Conservation of energy

Ans. The correct answer is d. Conservation of energy

Explanation: The first law of thermodynamics relates to the law of conservation of energy. It states that energy cannot be created or destroyed, but can only be converted from one form to another.

Ques. Entropy occurs due to

  1. Volumetric changes only
  2. Change in macroscopic variables
  3. Mass changes only
  4. Temperature only

Ans. The correct answer is b. Change in macroscopic variables

Explanation: Entropy is a measure of the disorder or randomness in a system. It is related to the number of possible microscopic configurations that a system can have. The more microscopic configurations a system can have, the higher its entropy.

Ques. An increase in enthalpy leads to an increase in

  1. Increase in internal energy
  2. Increase in pressure
  3. Increase in mass
  4. Increase in volume

Ans. The correct answer is a. Increase in internal energy

Explanation: An increase in temperature leads to an increase in molecular interactions. Using the equation from the first law of thermodynamics, internal energy increases as temperature rises. As a result, a rise in enthalpy results in an increase in internal energy.

Ques. During the process of photosynthesis, which reaction takes place?

  1. Redox reaction
  2. Exothermic reaction
  3. Endothermic reaction
  4. Chemical reaction

Ans. The correct answer is c. Endothermic reaction

Explanation: Photosynthesis occurs by absorbing heat and energy from the surroundings. An endothermic process is one in which the system absorbs heat from its surroundings. Plants absorb heat and energy from their surroundings during photosynthesis, making it an endothermic reaction.

Ques. The temperature developed during a fluid flow is due to

  1. Translational kinetic energy
  2. Increase in pressure
  3. Change in density
  4. Fluid level

Ans. The correct answer is a. Translational kinetic energy

Explanation: When there is a high rate of fluid flow, the molecules tend to collide with one another. The average translational kinetic energy of the particles increases in this state. The temperature developed due to this increase in kinetic energy is known as the Kinetic temperature.

Ques. Due to the increase in enthalpy, the internal energy

  1. Decreases
  2. Increases
  3. No change
  4. First decreases, then increases

Ans. The correct answer is b. Increases

Explanation: Since enthalpy is directly related to internal energy, an increase in enthalpy indicates an increase in internal energy.

Ques. What is the SI unit of enthalpy?

  1. kg/K
  2. Joule/K2
  3. Joule/kg
  4. kgK/Joule

Ans. The correct answer is c. Joule/kg

Explanation: Enthalpy is the measurement of energy in a thermodynamic system. It equals the product of volume and pressure plus internal energy. Thus, its unit is Joule/kg.

Ques. Isothermic process means that ΔU = 0

  1. True
  2. False

Ans. The correct answer is a. True

Explanation: Because internal energy is a function of temperature, ΔU = 0 indicates that the process will have a constant temperature. Therefore, it is isothermic.

Ques. Which of the following cannot determine the state of a thermodynamic system?

  1. Pressure and volume
  2. Volume and temperature
  3. Temperature and pressure
  4. Any one of pressure, volume, and temperature

Ans. The correct answer is d. Any one of pressure, volume, and temperature

Explanation: The state of a thermodynamic system cannot determined by a single variable (P, V, or T).


Short Answers Questions [2 Marks Questions]

Ques. What is thermodynamics?

Ans. Thermodynamics is a branch of physics that deals with the study of heat, work, and temperature in connection to energy, entropy, and the physical characteristics of matter and radiation.

Ques. What are the laws of thermodynamics?

Ans. The laws of thermodynamics are a collection of scientific laws that specify a group of physical quantities that characterize thermodynamic systems in thermodynamic equilibrium, such as temperature, energy, and entropy.

Ques. How many laws of thermodynamics are there?

Ans. There are four laws of thermodynamics

  • First law of thermodynamics
  • Second law of thermodynamics
  • Third law of thermodynamics
  • Zeroth law of thermodynamics

Ques. What is the first law of thermodynamics?

Ans. The first law of thermodynamics is a fundamental principle that governs the behavior of energy in thermodynamic systems. It states that the total energy within a closed system remains constant, even though it may transform from one form to another. In simple words, the heat transferred to a system is equivalent to the increase in its internal energy plus the work it performs on its surroundings.

Ques. What is the formula for the first law of thermodynamics?

Ans. The first law of thermodynamics can be mathematically represented as

ΔQ = ΔU + ΔW

Where

  • ΔQ is the heat energy given to the system
  • ΔU is the change in the internal energy of the system
  • ΔW is the work done by the system.

Ques. What is a reversible process?

Ans. A reversible process is one that involves a system and its surroundings and whose direction may be reversed by insignificant changes in some of the properties of the surroundings, such as pressure or temperature.

Ques. What is the formula for the internal energy of an ideal gas?

Ans. The internal energy of an ideal gas is the total kinetic energy of the gas molecules and is given by

U = nCVT

Where

  • n is the number of moles of gas
  • CV is the molar-specific heat at a constant volume
  • T is the temperature of the gas in Kelvin

Ques. What is an isolated system?

Ans. An isolated system is a thermodynamic system that does not exchange energy or matter with its surroundings. This means that neither energy nor matter can enter or leave the system.

Ques. What is the second law of thermodynamics?

Ans. The second law of thermodynamics is a physical law that is based on universal empirical observations about heat and energy interconversions. The law states simply that heat always flows spontaneously from hotter to colder regions of matter.

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Long Answers Questions [3 Marks Questions]

Ques. For a gaseous system find change in internal energy if the heat supplied to the system is 50 J and work done by the system is 16 J.

Ans. Given

  • Heat supplied to the system, ΔQ = 50 J
  • Work done by the system, ΔW = 16 J

Let ΔU be the change in internal energy, then according to the first law of thermodynamics, we have

ΔQ = ΔU + ΔW

⇒ ΔU = ΔQ - ΔW

On substituting the values, we get

ΔU = 50 - 16 = 34 J

Hence the change in internal energy of the system is 34 J.

Ques. If the amount of heat given to a system is 35 J and the amount of work done on the system is 15 J. Then, what is the change in the internal energy of the system?

Ans. Given

  • The amount of heat given to the system, ΔQ = 35 J
  • The amount of work done on the system, ΔW = - 15 J

Let ΔU be the change in internal energy, then according to the first law of thermodynamics, we have

ΔQ = ΔU + ΔW

⇒ ΔU = ΔQ - ΔW

On substituting the values, we get

ΔU = 35 - (-15) = 50 J

Hence the change in internal energy of the system is 34 J.

Ques. What are the limitations of the first law of thermodynamics?

Ans. The following are the limitations of the first law of thermodynamics

  • According to the law, when a system goes through a thermodynamic process, it must always maintain a perfect energy balance. The first law, on the other hand, fails to show the feasibility of the system's process or change of state. 
  • The first law does not explain why heat flows from the hot end to the cold end of a metallic rod when one end is heated but not the other, and vice versa. 
  • Only the amount of energy exchanged throughout this process can be measured by the first law.
  • The reverse process is not possible.

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 gives 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, the work done by the system on the surroundings is given by

ΔW = + 4.3 x 105 J

The heat released to the surroundings is, ΔQ = - 3.8 x 105 J

  1. Using the first law of thermodynamics,

ΔQ = ΔU + ΔW

⇒ ΔU = ΔQ - ΔW

On substituting the values, we get

Change in internal energy, ΔU = (- 3.8 x 105) - (+ 4.3 x 105 )

⇒ ΔU = - 8.1 x 105 J

  1. The amount of heat gives off by her while walking, ΔQ = - 1.2 x 105 J

Decrease in internal energy, ΔU = - 2.6 x 105 J

Using the first law of thermodynamics

ΔQ = ΔU + ΔW

⇒ ΔW = ΔQ - ΔU

On substituting the values, we get

ΔW = (- 1.2 x 105) - (- 2.6 x 105)

⇒ ΔW = 1.4 x 105 J

Ques. A gas is enclosed in a cylinder with a moveable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10-3 m3 changes to a final state at Pf = (1/32) x 105 Pa and Vf = 8 x 10-3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric process at volume Vf. What is the amount of heat supplied to the system in the two-step process?

Ans. Given

  • Initial pressure, Pi = 105 Pa
  • Final pressure, Pf = (1/32) x 105 Pa
  • Initial volume, Vi = 10-3 m3
  • The final volume, Vf = 8 x 10-3 m3 

Change in volume is given by

ΔV = Vf - Vi

⇒ ΔV = (8 x 10-3) - (10-3) = 7 x 10-3 m3

Work done by the system is given by

W = PiΔV

⇒ W = 105 x (7 x 10-3) = 700 J

Change in internal energy is given by

ΔU = nCvΔT

⇒ (3/2) [PfVf - PiVi] = (-900/8) J

From the first law of thermodynamics, we have

Heat supplied to the system, Q = ΔU + W

On substituting the values, we get

Q = (-900/8) + 700

⇒ Q = 588 J

Ques. An ideal gas is made to go through a cyclic thermodynamical process in four steps. The amount of energy involved is Q1 = 600 J, Q2 = -400 J, Q3 = -300 J, and Q4 = 200 J respectively. The corresponding work involved are W1 = 300 J, W2 = -200 J, W3 = -150 J, and W4. What is the value of W4?

Ans. From the first law of thermodynamics, we have

ΔQ = ΔU + ΔW

Where

  • ΔQ is the amount of heat
  • ΔU is the change in internal energy
  • ΔW is the amount of work done

For a cyclic process, change in internal energy is equal to zero i.e.

ΔU = 0

Therefore, the above equation becomes

ΔQ = ΔW …(i)

Now,

ΔQ = Q1 + Q2 + Q3 + Q4

On substituting the values, we get

ΔQ = 600 - 400 - 300 + 200 = 100 J

Also,

ΔW = W1 + W2 + W3 + W4

On substituting the values, we get

ΔW = 300 - 200 - 150 + W4 = -50 + W4

Using equation (i), we get

100 = -50 + W4

⇒ W4 = 150 J

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