12.1,2 The First law of thermal dynamics. Work done on the gas is, Also, we found that the heat transferred to the gas is, Kinetic and potential energy of the gas is called internal energy and the internal energy of an ideal gas is, From this definition, we can finally write down the law of conservation of energy of the gas. This is the first law of thermodynamics. Here we need to be careful about the signs of Q and W. Q>0 means Energy flows into the system. Q<0 means Energy flows out of the system. W>0 means Work is done on the system.(∆V<0) W<0 means Work is done by the system.(∆V>0)

Internal energy for a general system can be written as following, For an ideal gas, Cv = 3/2 R and for a general system, Where the degree of freedom is 3 for an ideal gas and 5 for a diatomic gas.

12.3 Thermal processes Before we talk thermal processes, we will introduce an ideal cylindrical engine and the PV diagram. This engine illustrates the energy flow of the system. We will use this engine to explain various thermal processes. Also, the PV diagram is very useful to use. For example, the area under the graph in a PV diagram is equal in magnitude to the work done on the system. Thermal processes we will study in this chapter are special cases of the first law of thermodynamics.

a. Isobaric Processes( ,Constant P) where Cp =5/2 R is the molar heat capacity at constant pressure. b. Adiabatic Process ( , Q = 0) This is the case in which the system is insulated. In an adiabatic process, the following is true. c. Isovolumetric Processes( , W=0 and constant V) This is a case the engine does not move, but the energy is transferred from the environment. d. Isothermal processes( ,∆U=0 and constant T) This can be an expansion of the gas in the cylinder while the temperature stays the same. Following equations are true in this case All the cases are summarized in the table shown in the textbook(see the next page).

Ch12 The laws of thermodynamics

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12.4 Heat Engine. A heat engine takes in energy by heat and partially convert it to other forms.

In general, a heat engine does work to an environment through a cyclic process.

1. energy is transferred by heat from a source at a high temperature.

2. Work is done by the engine.

3. Energy is expelled from the engine by heat to a source at the lower temperature.

4. Return to the original state.

Since initial and final states are the same, ∆U=0,

Here the Weng is the work done by the engine, which is –W.

Work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram.

Refrigerators and heat pumps are just heat engines operate in reverse.

The performance of refrigerators is,

The performance of heat pump is,

Both constants are dimensionless and usually

greater than 1.

12.4,5 The second law of thermodynamics and Entropy

According to the first law of thermodynamics,

For a cyclic process, so the maximum efficiency could be 1 if Qc=0, but this is impossible because of the second law of thermodynamics.

The second law of thermodynamics can be stated in various ways.

1. No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the performance of an equal amount of work.

2. If two systems are in thermal contact, net thermal energy transfers spontaneously by heat from the hotter system to the colder system.

3. The entropy of the Universe increases in all natural processes(irreversible processes).

Entropy is a measure of disorder. Mathematically, it is defined as

Where Qr is the energy absorbed or expelled during a reversible, constant temperature process between two equilibrium states.

This value represents the number of possible configurations in a system. (recall the example

we talk about in the class.)

We can use this equation to calculate the entropy change by a heat engine.

Reversible and irreversible processes are defined as,

“In a reversible process, every state along the path is an equilibrium state, so the system can return to its initial condition by going along the same path in the reverse direction”

“If a process does not satisfy above requirement, it is irreversible.”

Most natural processes are known to be irreversible.

The second law is different from other laws we studied in this class.

1. This law arises from the statistic analysis of a many-body system, so it is really a statement of what is most probable rather than of what must be.

2. It is not symmetric in time. Other laws do not change if the time is reversed, but the second law indicates the direction of the time.

Please watch this youtube video for more explanations about the entropy https://youtu.be/vE82PDJB8ow

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12.5 Carnot engine and the third law of thermodynamics.

The most efficient engine undergoes an ideal, reversible cycle, which is called Carnot cycle.

Thus, Carnot engine is the most efficient engine.

In order for the engine to operate in an ideal, reversible cycle, it must satisfy the following conditions.

1. Energy is transferred only in the equilibrium state.(Isothermal process)

2. When the system is not in equilibrium with the environment, energy is not transferred.(adiabatic)

Then the Carnot cycle can be shown in the figure below.

1. The process A to B is an isothermal expansion at Th. During the process, the gas absorbs energy Qh and does work WAB.

2. In the process B to C, the system is insulated and the gas expands adiabatically. During the process, the temperature falls from Th to Tc and the gas does work WBC.

3. In the process C to D, the gas compressed

isothermally at temperature Tc. During the process, the gas expels energy Qc to the reservoir and the work done on the gas is WCD.

4. In the final process from D to A, the system is compressed adiabatically. During the process the temperature increases from Tc to Th and the work done on the gas is WDA.

The efficiency of the Carnot engine is,

From this equation, it seems the efficiency can be 1 if Tc=0 K. However, Tc can not be zero according to the third law of thermodynamics.

“it is impossible to lower the temperature of a system to absolute zero in a finite number of steps”

This is the reason why the efficiency of the Carnot cycle is always less than 1.

Useful information about thermal processes.

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11

A monatomic ideal gas has a temperature T = 300K and a constant volume of 1.50 L. If there are 5.00 mole of gas, (a) how much thermal energy must be added in order to raise the temperature of the gas to 380K? (b) Calculate the change in pressure of the gas, ∆P.

A heat engine contains an ideal monatomic gas confined to a cylinder by a movable piston. The gas starts at A, where T = 300 K. The process B to C is an isothermal expansion. (a)Find the number n of moles of gas and the temperature at B. (b) Find ∆U, Q and W for the isovolumetric process A to B. (c) Repeat for the isothermal process C to A (d) Repeat for the isobaric process C to A. (e) find the thermal energy Qh transferred into the system, the thermal energy rejected, Qc the thermal efficiency, and net work on the environment performed by the engine.

A 0.002 m3 container of left over soup at a temperature of 323 K is placed in a refrigerator. Assume the specific heat of the soup is 4190 J/kg K and the density is 1.25 x

103 kg/m3. The refrigerator cools the soup to 283 K. If the COP of the refrigerator is 5.00, find the energy needed to cool the soup.

A steam engine has a boiler that operates at 5.00 x 102 K. The energy from the boiler changes water to steam, which drives the piston. The temperature of the exhaust is that of the outside air, 3.00 x 102 K. (a) What is the engine’s efficiency if it’s an ideal engine? (b) If the 3.50 x 103 J of energy is supplied from the boiler, find the energy transferred to the cold reservoir and the work done by the engine on its environment.

Find the change in entropy of 3.00 x 102 g of lead when it melts at 327°C. Lead has a latent heat of fusion of 2.45 x 104 J/kg.

A heat engine contains an ideal monatomic gas confined to a cylinder by a movable piston. The gas starts at A, where T = 300 K. The process B to C is an isothermal expansion. (a)Find the number n of moles of gas and the temperature at B. (b) Find ∆U, Q and W for the isovolumetric process A to B. (c) Repeat for the process C to A (d) Repeat for the isobaric process C to A. (e) find the thermal energy Qh transferred into the system, the thermal energy rejected, Qc the thermal efficiency, and net work on the environment performed by the engine. L5

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