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10.1Internal energy
This chapter deals with how we apply the law of conservation of energy to thermodynamics. To do so, we must first discuss the concept of internal energy, how we calculate work done in thermodynamics and discuss various processes using pressureâvolume diagrams and equations.
In ChapterÂ 7 we defined the internal energy of a gas as the total random kinetic energy of the particles of the gas plus the potential energy associated with the intermolecular forces. If the gas is ideal, the intermolecular forces are assumed to be strictly zero, and the internal energy of the gas is just the total random kinetic energy of the particles of the gas. We have seen that the average kinetic energy of the particles is given by:
EÂ¯K=32kBT
where kB=RNA=1.38Ã10â23Â JÂ Kâ1 is the Boltzmann constant.
It follows that the internal energy U of an ideal gas with N particles is given by NEÂ¯K; that is:
U=32NkBT
This formula shows that the internal energy of an ideal gas is directly proportional to the (kelvin) temperature.
Since kB=RNA and PV = nRT, we also have that:
U=32nRT=32PV
where n=NNA is the number of moles.
EXAM TIP
You must know the equivalent ways of expressing internal energy.
EXAM TIP
Throughout this chapter we deal with monatomic gases.
The change in internal energy is thus given by the formulae:
ÎU=32NkBÎT=32nRÎT=32PÎVîîîîîîifÂ P=const=32VÎPîîîîîîifÂ V=Â constÂ 
CHECK YOURSELF 1
A fixed quantity of an ideal gas at pressure 6.0 Ã 105 Pa and volume 3.0 Ã 10â3 m3 undergoes a change to pressure 9.0 Ã 105 Pa and volume 2.0 Ã 10â3 m3. What is the change in the internal energy of the gas?
Systems and processes
In thermodynamics we often deal with systems; this means the complete set of objects under consideration. Thus, a gas in a container is a system, as is a certain mass of ice in a glass. What is not in the system is the surroundings of the system. Heat can enter or leave a system depending on its temperature relative to that of the surroundings, as shown in Figure 10.1.
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Figure 10.1: A system and its surroundings.

In a closed system mass cannot enter or leave the system. ln an isolated system, no energy or mass can enter or leave the system. If all the parameters defining the system are given, we speak of the system being in a particular state. For example, a fixed quantity of an ideal gas is specified if its pressure, volume and temperature are specified. Any process that changes the state of a system is called a thermodynamic process. Thus, heating a gas may result in changed pressure, temperature or volume and is thus a thermodynamic process. Doing work on the gas by compressing it is another thermodynamic process.
Internal energy is a property of the particular state of the system under consideration, and for this reason internal energy is called a state function. Thus, if two equal quantities of an ideal gas originally in different states are brought to the same state (i.e. same pressure, volume and temperature), they will have the same internal energy irrespective of what the original state was and how the gas was brought to that final state. By contrast, heat and work are not state functions. Heat and work are related to changes in the state of the system, not to the state itself.
Process
Description
Equation
Isothermal
Temperature stays constant
P1V1=P2V2
Isobaric
Pressure stays constant
V1T1=V2T2
Isovolumetric
Volume stays constant
P1T1=P2T2
Adiabatic
No heat enters or leaves the system
P1V5/31=P2V5/32
Table 10.1: A few thermodynamic processes.
In this course we will be mainly interested in four different types of thermodynamic processes. These are defined in Table 10.1. We have seen all the equations before except that for the adiabatic process which we will discuss shortly.
Pressureâvolume diagrams
Changes in a gas may be conveniently shown on pressureâvolume diagrams. Examples of isovolumetric (red) and isobaric (blue) processes are shown in Figure 10.2.

Figure 10.2: Isobaric and isovolumetric processes

The adiabatic equation
In an adiabatic process no heat enters or leaves the system. It can be shown that if the state of a fixed quantity of an ideal gas is changed adiabatically from pressure P1 and volume V1 to pressure P2 and volume V2 we have that
P1V531=P2V532
i.e. during an adiabatic process
PV53=Â constantÂ =c1
This means that in an adiabatic change the temperature also changes. From the ideal gas law we have that PVT=Â constantÂ =c2. Solving this for pressure we get P=c2TV, and substituting this in the adiabatic law (c2TV)V53=c1; that is:
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TV23=Â constant or T1V231=T2V232.
So, when a gas is compressed adiabatically, the temperature increases. An everyday example of this is pumping a bicycle tire: the valve gets very hot. This will be explained shortly.
WORKED EXAMPLE 10.1
An ideal gas expands adiabatically from a state with pressure 5.00 Ã 105 Pa, volume 2.20 Ã 10â3 m3 and temperature 485 K to a new volume of 3.80 Ã 10â3 m3. Calculate the new pressure and temperature of the gas.
Answer
From P1V531=P2V532 we find:
5.00Ã105Ã(2.20Ã10â3)53=P2Ã(3.80Ã10â3)53
P2 = 5.00Ã105Ã(2.20Ã10â3)53(3.80Ã10â3)53 = 5.00Ã105Ã(2.20Ã10â33.80Ã10â3)33 = 5.00Ã105Ã(2.203.80)53 = 2.01Ã105Â Pa
To find the new temperature we may use T1V231=T2V232, but it is simpler to use the ideal gas law and write P1V1T1=P2V2T2
so that:
5.00Ã105Ã2.20Ã10â3485=2.01Ã105Ã3.80Ã10â3T
T2=2.01Ã105Ã3.80Ã10â3Ã4855.00Ã105Ã2.20Ã10â3=337Â K
An adiabatic expansion on a pressure volume diagram is steeper than an isothermal. Figure 10.3a shows an isothermal (blue) and an adiabatic (red) expansion of the same gas from a common state. Figure 10.3b shows an isothermal and adiabatic compression of the same gas from a common state.

Figure 10.3: a Isothermal and adiabatic expansions form the same state. b Isothermal and adiabatic compressions from the same state. In both cases the adiabatic curve is steeper.

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Isothermal processes are slow (to allow time for the temperature to equalize and stay constant) while adiabatic processes are fast (so there is no time for heat to be exchanged).
We know that temperature increases in an adiabatic compression. We will be able to understand this when we learn about the first law of thermodynamics. But we can also understand it in terms of molecular motion: imagine a gas in a container with a piston; the gas is compressed so the piston moves inwards. The process is adiabatic so the piston has to move fast. Molecules colliding with the inward moving piston will bounce back with increased speed. The average kinetic energy of the molecules will increase and so the temperature has to increase.
CHECK YOURSELF 2
Consider Figure 10.3b and look at the point on each curve corresponding to a volume of 1.0 Ã 10â3 m3. Explain using the graph why the point on the adiabatic has a higher temperature.
Work done on or by a gas
Consider a quantity of a gas in a container with a frictionless, movable piston, Figure 10.4. If the piston is to stay in place a force from the outside has to be applied to counterbalance the force due to the pressure of the gas. Let us now compress the gas by pushing the piston in with a very slightly greater force.

Figure 10.4: When the piston is pushed in by a small amount, work is being done on the gas.

If the pressure in the gas initially is P, and the cross-sectional area of the piston is A, then the force with which one must push is PA. If the pressure stays constant, the force is constant and so we may use the result from mechanics that work is force times distance moved in the direction of the force. The piston moves a distance Îs, so the work done is
W=FÎs=PAÎs
But A Îs is change in the volume of the gas, ÎV. The same result holds if, instead, the piston is moved outwards by the gas as it expands.
KEY POINT
The work done when the volume of the gas is changed by ÎV at constant pressure is W = P Î V.
Figure 10.5 shows that the work done is the area under the isobaric curve; the area is a rectangle of base ÎV and height P. The work done in this case is
W=1.5Ã105Ã(1.6Ã10â3â0.4Ã10â3)=180Â J.

Figure 10.5: The area under the graph in a Â­pressure-volume diagram gives the work done.

What if the pressure changes? Then the force is not constant and we have to do what we did in mechanics: the work done is the area under the curve in a forceâdistance graph. Here the corresponding result is the area under the pressureâvolume graph.
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KEY POINT
The work done when the gas expands by an arbitrary amount is the area under the graph in the pressureâvolume diagram.
In Figure 10.6 we have to calculate the area of the trapezium:
W=(1.4+2.6)Ã1052Ã1.2Ã10â3=240Â J

Figure 10.6: The work done is found from the area under the curve in the pressureâvolume diagram.

The pressureâvolume diagrams in Figure 10.7 show an ideal gas that starts expanding from state A to state B and is then compressed back to state A along a different path. The gas does work in expanding from A to B, but work is done on the gas in compressing it back to A. The net work done therefore is the area of the loop.

Figure 10.7: For a closed loop in a pressureâvolume diagram, the work done is the area of the loop.

CHECK YOURSELF 3
An ideal gas has a constant volume 5.0 Ã 10â3 m3.It is heated so that the pressure increases from 2.0 Ã 105 Pa to 6.0 Ã 105 Pa. What is the work done?
WORKED EXAMPLE 10.2
A gas is compressed at constant pressure 2.00 Ã 105 Pa from a volume of 2.00 m3 to a volume of 1.75 m3.
aFind the work done.
The temperature was initially 240.0 Â°C.
bCalculate the final temperature of the gas in Â°C.
Answer
aSince the compression takes place under constant pressure, the work done is
PÃÂ change in volumeÂ =2.00Ã105Â PaÃ0.25Â m3=5.0Ã104Â J
bThe final temperature is found from
VT=Â constantÂ 
that is,
2.00513=1.75T
giving
T=449Â K=176Â°C.
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TEST YOUR UNDERSTANDING
1Why are isothermal processes slow and adiabatic processes fast?
2Using ideas on molecular motion explain why in an adiabatic expansion of an ideal gas the temperature drops.
3An ideal gas expands adiabatically from a state with pressure 8.1 Ã 105 Pa and volume 2.5 Ã 10â3 m3 to a state of volume 4.6 Ã 10â3 m3. Calculate the new pressure of the gas.
4An ideal gas expands adiabatically from a state with volume 2.8 Ã 10â3 m3 and temperature 560 K to a state of volume 4.8 Ã 10â3 m3. Calculate the new temperature of the gas.
5A gas expands at a constant pressure of 5.4 Ã 105 Pa from a volume of 3.6 Ã 10â3 m3 to a volume of 4.3 Ã 10â3 m3. Calculate the work done by the gas.
6A gas is compressed isothermally from a given state to final volume V. An amount of work W is done on the gas. If instead the gas was compressed adiabatically from the same state to a state of volume V, would the work done on the gas be equal to, greater than or less than W?
7An ideal gas expands isothermally from pressure P and volume V to volume 2V. Sketch this change on a pressureâvolume diagram. An equal quantity of an ideal gas at pressure P and volume V expands adiabatically to a volume 2V. Sketch this change on the same axes. Determine in which case the work done by the gas larger.
