Question

4. An ideal gas in a moveable piston is allowed to reversibly expand by slowly heating at constant pressure.

a. (10 pts) Derive the equation for the work done (w reversible) as a function of the initial and final temperatures.

b. (10 pts) Derive the equation for the change in heat in the system (q).

c. (10 pts) Derive the equation for the change in entropy.

Answer 1

A)Derive the equation for the work done (w reversible) as a function of the initial and final temperatures

To understand the relationship between work and heat, we need to understand a third, linking factor: the change in internal energy. Energy cannot be created nor destroyed, but it can be converted or transferred. Internal energy refers to the all the energy within a given system, including the kinetic energy of molecules and the energy stored in all of the chemical bonds between molecules. With the interactions of heat, work and internal energy, there are energy transfers and conversions every time a change is made upon a system. However, no net energy is created or lost during these transfers.

Definition: 1st Law of Thermodynamics

The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as

ΔU=q+w ---------(1)

with

• ΔU is the total change in internal energy of a system,
• q is the heat exchanged between a system and its surroundings, and
• w is the work done by or on the system.

Work is also equal to the negative external pressure on the system multiplied by the change in volume:

w=−pΔV ---------(2)

where P is the external pressure on the system, and ΔV is the change in volume. This is specifically called "pressure-volume" work.

The internal energy of a system would decrease if the system gives off heat or does work. Therefore, internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done onto a system. Any work or heat that goes into or out of a system changes the internal energy. However, since energy is never created nor destroyed (thus, the first law of thermodynamics), the change in internal energy always equals zero. If energy is lost by the system, then it is absorbed by the surroundings. If energy is absorbed into a system, then that energy was released by the surroundings:

ΔUsystem=−ΔUsurroundings

where  ΔUsystem is the total internal energy in a system, and ΔUsurroundingsis the total energy of the surroundings.

Table:1

Process	Sign of heat (q)	Sign of Work (w)
Work done by the system	N/A	-
Work done onto the system	N/A	+
Heat released from the system- exothermic (absorbed by surroundings)	-	N/A

Table:2

The Process	Internal Energy	Heat Transfer of Thermal Energy (q)	Work (w=-PΔV)	Example
q=0 Adiabatic	+	0	+	Isolated system in which heat does not enter or leave similar to styrofoam
ΔV=0 Constant Volume	+	+	0	A hard, pressure isolated system like a bomb calorimeter
Constant Pressure	+ or -	enthalpy (ΔH)	-PΔV	Most processes occur are constant external pressure
ΔT=0 Isothermal	0	+	-	There is no change in temperature like in a temperature bath

B) Derive the equation for the change in heat in the system (q)

When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal to the change in enthalpy. Enthalpy (H) is the sum of the internal energy (U) and the

product of pressure and volume (PV) given by the equation:

H=U+PV

When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal to the change in enthalpy. Enthalpy is a state function which depends entirely on the state functions T, P and U. Enthalpy is usually expressed as the change in enthalpy (ΔH) for a process between initial and final states:

ΔH=ΔU+ΔPV

If temperature and pressure remain constant through the process and the work is limited to pressure-volume work, then the enthalpy change is given by the equation:

ΔH=ΔU+PΔV

Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined by the equation:

ΔH=qΔH=q

By looking at whether q is exothermic or endothermic we can determine a relationship between ΔH and q. If the reaction absorbs heat it is endothermic meaning the reaction consumes heat from the surroundings so q>0 (positive). Therefore, at constant temperature and pressure, by the equation above, if q is positive then ΔH is also positive. And the same goes for if the reaction releases heat, then it is exothermic, meaning the system gives off heat to its surroundings, so q<0 (negative). If q is negative, then ΔH will also be negative.

Enthalpy Change Accompanying a Change in State

When a liquid vaporizes the liquid must absorb heat from its surroundings to replace the energy taken by the vaporizing molecules in order for the temperature to remain constant. This heat required to vaporize the liquid is called enthalpy of vaporization (or heat of vaporization). For example, the vaporization of one mole of water the enthalpy is given as:

ΔH = 44.0 kJ at 298 K

When a solid melts, the required energy is similarly called enthalpy of fusion (or heat of fusion). For example, one mole of ice the enthalpy is given as:

ΔH = 6.01 kJ at 273.15 K

ΔH=ΔU+pΔV -----------(1)

Enthalpy can also be expressed as a molar enthalpy, ΔHm, by dividing the enthalpy or change in enthalpy by the number of moles. Enthalpy is a state function. This implies that when a system changes from one state to another, the change in enthalpy is independent of the path between two states of a system.

If there is no non-expansion work on the system and the pressure is still constant, then the change in enthalpy will equal the heat consumed or released by the system (q).

ΔH=q ----------(2)

This relationship can help to determine whether a reaction is endothermic or exothermic. At constant pressure, an endothermic reaction is when heat is absorbed. This means that the system consumes heat from the surroundings, so qq is greater than zero. Therefore according to the second equation, the ΔH will also be greater than zero. On the other hand, an exothermic reaction at constant pressure is when heat is released. This implies that the system gives off heat to the surroundings, so q is less than zero. Furthermore, ΔH will be less than zero.

Effect of Temperature on Enthalpy

When the temperature increases, the amount of molecular interactions also increases. When the number of interactions increase, then the internal energy of the system rises. According to the first equation given, if the internal energy (U) increases then the ΔH increases as temperature rises. We can use the equation for heat capacity and Equation 2 to derive this relationship.

C=qΔT ----------(3)

Under constant pressure, substitute Equation 2 into equation 3:

Cp=(ΔHΔT)P----------(4)

where the subscript P indicates the derivative is done under constant pressure.

The Enthalpy of Phase Transition

Enthalpy can be represented as the standard enthalpy, ΔHoΔHo. This is the enthalpy of a substance at standard state. The standard state is defined as the pure substance held constant at 1 bar of pressure. Phase transitions, such as ice to liquid water, require or absorb a particular amount of standard enthalpy:

• Standard Enthalpy of Vaporization (ΔHovapΔHvapo) is the energy that must be supplied as heat at constant pressure per mole of molecules vaporized (liquid to gas).

• Standard Enthalpy of Fusion (ΔHofus) is the energy that must be supplied as heat at constant pressure per mole of molecules melted (solid to liquid).

• Standard Enthalpy of Sublimation (ΔHosub) is the energy that must be supplied as heat at constant pressure per mole of molecules converted to vapor from a solid.

ΔHosub=ΔHofus+ΔHovap

The enthalpy of condensation is the reverse of the enthalpy of vaporization and the enthalpy of freezing is the reverse of the enthalpy of fusion. The enthalpy change of a reverse phase transition is the negative of the enthalpy change of the forward phase transition. Also the enthalpy change of a complete process is the sum of the enthalpy changes for each of the phase transitions incorporated in the process.

C) Derive the equation for the change in entropy

Entropy Changes in an Ideal Gas

Many aerospace applications involve flow of gases (e.g., air) and we thus examine the entropy relations for ideal gas behavior. The starting point is form (a) of the combined first and second law,

For an ideal gas, . Thus

Using the equation of state for an ideal gas ( ), we can write the entropy change as an expression with only exact differentials:

(5..2)

We can think of Equation (5.2) as relating the fractional change in temperature to the fractional change of volume, with scale factors and ; if the volume increases without a proportionate decrease in temperature (as in the case of an adiabatic free expansion), then increases. Integrating Equation (5.2) between two states ``1'' and ``2'':

For a perfect gas with constant specific heats

In non-dimensional form (using )

(5..3)

Equation 5.3 is in terms of specific quantities. For moles of gas,

This expression gives entropy change in terms of temperature and volume. We can develop an alternative form in terms of pressure and volume, which allows us to examine an assumption we have used. The ideal gas equation of state can be written as

Taking differentials of both sides yields

Using the above equation in Eq. (5.2), and making use of the relations ; , we find

or

Integrating between two states 1 and 2

(5..4)

Using both sides of (5.4) as exponents we obtain

(5..5)

Equation (5.5) describes a general process. For the specific situation in which , i.e., the entropy is constant, we recover the expression . It was stated that this expression applied to a reversible, adiabatic process. We now see, through use of the second law, a deeper meaning to the expression, and to the concept of a reversible adiabatic process, in that both are characteristics of a constant entropy, orisentropic, process.