Notes : First Law of Thermodynamics with Equations and Processes

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First Law of Thermodynamics

The First Law of Thermodynamics is the law of conservation of energy applied to thermodynamic systems. It states that the heat supplied to a system is equal to the increase in its internal energy plus the work done by the system.

Mathematical Equation

$$\Delta Q = \Delta U + \Delta W$$

Where:

  • ΔQ = Heat supplied to the system
  • ΔU = Change in internal energy
  • ΔW = Work done by the system

For expansion against constant pressure:

$$\Delta W = P\Delta V$$

Physical Significance of First Law of Thermodynamics

The First Law explains the relationship between heat, work and internal energy.

  • Heat supplied may increase internal energy.
  • Heat supplied may be converted into work.
  • Total energy remains conserved.

First Law in Different Thermodynamic Processes

1. Isothermal Process

Temperature remains constant.

$$\Delta T = 0$$

For an ideal gas:

$$\Delta U = 0$$

Therefore:

$$\Delta Q = \Delta W$$

2. Isochoric Process

Volume remains constant.

$$\Delta V = 0$$

$$\Delta W = P\Delta V = 0$$

Therefore:

$$\Delta Q = \Delta U$$

3. Isobaric Process

Pressure remains constant.

$$\Delta P = 0$$

$$\Delta W = P\Delta V$$

Therefore:

$$\Delta Q = \Delta U + P\Delta V$$

4. Adiabatic Process

No heat exchange occurs.

$$\Delta Q = 0$$

Therefore:

$$\Delta U = -\Delta W$$

5. Cyclic Process

The system returns to its initial state.

$$\Delta U = 0$$

Therefore:

$$\Delta Q = \Delta W$$

6. Isolated System

No heat exchange and no work done.

$$\Delta Q = 0$$

$$\Delta W = 0$$

Therefore:

$$\Delta U = 0$$

Example: Vaporization of 1 g Water

This example demonstrates how the First Law of Thermodynamics applies when 1 gram of water changes into steam.

Given Data

  • Mass of water = 1 g
  • Latent heat of vaporization = 2256 J g⁻¹
  • Atmospheric pressure = 1.013 × 10⁵ Pa
  • Initial volume = 1 cm³
  • Final volume = 1671 cm³

Step 1: Heat Supplied

$$\Delta Q = mL$$

$$\Delta Q = 1 \times 2256 = 2256\,J$$

Step 2: Work Done by Expanding Steam

$$\Delta V = V_g - V_l$$

$$\Delta V = 1.67 \times 10^{-3}\,m^3$$

$$\Delta W = P\Delta V$$

$$\Delta W \approx 169.2\,J$$

Step 3: Change in Internal Energy

$$\Delta U = \Delta Q - \Delta W$$

$$\Delta U = 2256 - 169.2$$

$$\Delta U = 2086.8\,J$$

Conclusion

Out of 2256 J of heat supplied, about 169.2 J is used to perform external work and 2086.8 J increases the internal energy of the system.

Solved Numerical

Question: A system absorbs 500 J of heat and performs 200 J of work. Calculate the change in internal energy.

Solution:

$$\Delta U = \Delta Q - \Delta W$$

$$\Delta U = 500 - 200$$

$$\Delta U = 300\,J$$

Limitations of First Law of Thermodynamics

  • It does not indicate the direction of heat flow.
  • It cannot explain spontaneous processes.
  • It does not determine whether a process is possible or impossible.
  • It does not place any restriction on heat-to-work conversion.

Frequently Asked Questions (FAQs)

What is the First Law of Thermodynamics?

The heat supplied to a system equals the increase in internal energy plus the work done by the system.

What is the equation of the First Law?

$$\Delta Q = \Delta U + \Delta W$$

Why is the First Law important?

It expresses the conservation of energy for thermodynamic systems.

MCQs on First Law of Thermodynamics

1. The First Law of Thermodynamics is based on:

A) Conservation of Momentum

B) Conservation of Charge

C) Conservation of Energy

D) Conservation of Mass

Answer: C) Conservation of Energy

2. In an isochoric process, work done is:

A) Positive

B) Negative

C) Zero

D) Maximum

Answer: C) Zero

3. In an adiabatic process:

A) ΔQ > 0

B) ΔQ < 0

C) ΔQ = 0

D) ΔU = 0

Answer: C) ΔQ = 0

4. In a cyclic process, change in internal energy is:

A) Positive

B) Negative

C) Maximum

D) Zero

Answer: D) Zero

5. The SI unit of internal energy is:

A) Watt

B) Joule

C) Pascal

D) Kelvin

Answer: B) Joule

Key Points to Remember

  • The First Law of Thermodynamics is based on conservation of energy.
  • Internal energy is a state function.
  • Heat and work are path functions.
  • For an isolated system, internal energy remains constant.
  • The law relates heat, work and internal energy.

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