# Thermal and Statistical Physics Essay Example

• Category:
Physics
• Document type:
Math Problem
• Level:
Masters
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715

13THERMAL AND STATISTICAL PHYSICS

Thermal and Statistical Physics

Thermal and Statistical Physics

Question 1: Efficiency of a cannot cycle

The area under the curve is as below:

i

This equation gives the amount of energy transferred in the process

1. If the process moves towards the less entropy, then heat will be removed from the system.

2. If the process moves towards the great entropy, then heat will be absorbed in the system.

From the TS diagram for a reversible, the amount of work done over a cyclic process is:

(ii)

Evaluating equation (ii) above:

Thus the amount of energy transferred from the hot reservoir to the cold reservoir will be:

and amount of energy transferred between the system and the cold reservoir will be:

Therefore the efficiency,

Is the maximum system entropy

Is the minimum system entropy

The amount of heat entering the system

The amount of heat leaving the system

The absolute temperature of the hot reservoir

The absolute temperature of the cold reservoir

Question 2

1. For a closed system

Due to charge in temperature and volume, the internal energy charges as:

Integrating the equation above:

1. The change in internal energy of a closed system

Because of compression or expansion of the system, the equation becomes as below replacing the work in the system.

Taking F as a function of T and V, the heat capacity is as below:

1. From the law of thermodynamics, for a reversible process:

Differentiating the internal energy equation

Hence,

Therefore

1. Heat capacity with constant pressure

At constant pressure:

Simplifying

Therefore at constant pressure:

So

1. For a constant temperature and volume

Change in heat capacity with constant volume is given as:

The change in internal energy of a closed system

Because of compression or expansion of the system, the equation becomes as below replacing the work in the system:

Taking F as a function of T and V, the heat capacity is as below:

Therefore

Question 3: Heat transfer

1. Heat transferred,

Where: Q it the amount of heat transferred.

M is the mass

is the heat capacity of the atmosphere

Is the temperature change

1. Work done

Where: W = work done

R is the ideal gas constant

P1 is the pressure at the beginning

2 is the final pressure

T it is the temperature

-19703.72 J energy is lost in the system

1. Change in internal energy

Therefore

Question 4

1. The change in internal energy of a closed system

Because of compression or expansion of the system, the equation becomes as below replacing the work in the system:

Therefore at constant volume and entropy

1. At constant entropy

Taking entropy at constant volume and temperature:

This becomes,

1. At constant entropy and pressure

Therefore diving the and at constant pressure and entropy

Therefore

But

And

Therefore,

So,

Question 5

1. Phase diagram of a fluid

1. Phase transitions

1. Two properties of the system that change

During the phase transition, the properties that change are pressure and volume.

The two properties that remain constant are temperature and mass.

1. Clapeyron’s Equation

The slope of the curve follows:

During the phase change, the temperature is constant:

From the Maxwell’s relation:

Taking an integral from one phase to another:

For a closed system and using the first law of thermodynamics:

Using the fact that temperature and pressure are constant, we get:

Therefore,

Substituting in the equation of pressure change:

Therefore, the equation becomes:

1. Clapeyron’s equation of sublimation

From the equation,

Integrating the equation,

Therefore for a liquid vapor boundary:

Given pressure and temperature are constant, therefore:

Question 6

1. Given the vapor pressure for liquid and solid and solving the equations simultaneously:

i

ii

Solving (i) and (ii) simultaneously

1. Latent heat for triple points

At the triple points

Latent heat

Question 7

1. Maximum inversion temperature of Helium

From the equation

Maximum inversion temperature is at

For

0 =

Therefore, the maximum inversion temperature of Helium is

1. Maximum inversion pressure is at maximum inversion temperature.

At 24 K, the inversion curve is at maximum therefore,

References

## Balmer, R. T. (2010). Modern engineering thermodynamics. New York: Elsivier.

Clausius, R. (1879). The mechanical theory of heat.London: Macmillan & Co.

Kittel, C. (1986). Introduction to solid state physics (6th edition). New York: John Wiley.

Stanley, H. E. (1971). Introduction to phase transitions and critical phenomena. New York:

Oxford University Press.