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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 the sum of the random kinetic energy of particles and the intermolecular potential energy in a system, how we calculate work done in thermodynamics and discuss various processes using pressure–volume diagrams and equations.

In Chapter 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:

E¯K=3 2 k B T

where k B=R N A=1.38×10−23 J K−1 is the Boltzmann constant.

It follows that the internal energy U of an ideal gas with N particles is given by N E¯K; that is:

U=3 2 N k B T

This formula shows that the internal energy of an ideal gas is directly proportional to the (kelvin) temperature.

Since k B=R N A and PV = nRT, we also have that:

U=3 2 n R T=3 2 P V

where n=N N A 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:

Δ U=3 2 N k B Δ T=3 2 n R Δ T=3 2 P Δ V⏟if P=const=3 2 V Δ P⏟if V= const

CHECK YOURSELF 1

A fixed quantity of an ideal gas at pressure 6.0 × 105 Pa and volume 3.0 × 10−3 m3 undergoes a change to pressure 9.0 × 105 Pa and volume 2.0 × 10−3 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 a system that can transfer energy, but not matter (mass), into or from its surroundings mass cannot enter or leave the system. ln an isolated system a system where no energy or matter (mass) can be transferred into or from its surroundings, 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 a thermodynamic process in which temperature stays constant | Temperature stays constant | P 1 V 1=P 2 V 2
Isobaric a thermodynamic process in which pressure stays constant | Pressure stays constant | V 1 T 1=V 2 T 2
Isovolumetric a thermodynamic process in which volume stays constant | Volume stays constant | P 1 T 1=P 2 T 2
Adiabatic a thermodynamic process in which no heat enters or leaves the system | No heat enters or leaves the system | P 1 V 1 5/3=P 2 V 2 5/3

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.

Pressure–volume diagrams

Changes in a gas may be conveniently shown on pressure–volume 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 P 1 and volume V 1 to pressure P 2 and volume V 2 we have that

P 1 V 1 5 3=P 2 V 2 5 3

i.e. during an adiabatic process

P V 5 3= constant =c 1

This means that in an adiabatic change the temperature also changes. From the ideal gas law we have that P V T= constant =c 2. Solving this for pressure we get P=c 2 T V, and substituting this in the adiabatic law (c 2 T V)V 5 3=c 1; that is:

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T V 2 3= constant or T 1 V 1 2 3=T 2 V 2 2 3.

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 × 105 Pa, volume 2.20 × 10–3 m3 and temperature 485 K to a new volume of 3.80 × 10–3 m3. Calculate the new pressure and temperature of the gas.

Answer

From P 1 V 1 5 3=P 2 V 2 5 3 we find:

5.00×10 5×(2.20×10−3)5 3=P 2×(3.80×10−3)5 3

P 2 = 5.00×10 5×(2.20×10−3)5 3(3.80×10−3)5 3 = 5.00×10 5×(2.20×10−3 3.80×10−3)3 3 = 5.00×10 5×(2.20 3.80)5 3 = 2.01×10 5 Pa

To find the new temperature we may use T 1 V 1 2 3=T 2 V 2 2 3, but it is simpler to use the ideal gas law and write P 1 V 1 T 1=P 2 V 2 T 2

so that:

5.00×10 5×2.20×10−3 485=2.01×10 5×3.80×10−3 T

T 2=2.01×10 5×3.80×10−3×485 5.00×10 5×2.20×10−3=337 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 × 10−3 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 Δs, so the work done is

W=F Δ s=P A Δ s

But A Δs is change in the volume of the gas, ΔV. 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 ΔV at constant pressure is W = P Δ V.

Figure 10.5 shows that the work done is the area under the isobaric curve; the area is a rectangle of base ΔV and height P. The work done in this case is

W=1.5×10 5×(1.6×10−3−0.4×10−3)=180 J.

Figure 10.5: The area under the graph in a pressure-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 force–distance graph. Here the corresponding result is the area under the pressure–volume 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 pressure–volume diagram.

In Figure 10.6 we have to calculate the area of the trapezium:

W=(1.4+2.6)×10 5 2×1.2×10−3=240 J

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

The pressure–volume 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 pressure–volume diagram, the work done is the area of the loop.

CHECK YOURSELF 3

An ideal gas has a constant volume 5.0 × 10−3 m3.It is heated so that the pressure increases from 2.0 × 105 Pa to 6.0 × 105 Pa. What is the work done?

Worked example 10.2

A gas is compressed at constant pressure 2.00 × 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 °C.

bCalculate the final temperature of the gas in °C.

Answer

aSince the compression takes place under constant pressure, the work done is

P× change in volume =2.00×10 5 Pa×0.25 m 3=5.0×10 4 J

bThe final temperature is found from

V T= constant

that is,

2.00 513=1.75 T

giving

T=449 K=176°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 × 105 Pa and volume 2.5 × 10–3 m3 to a state of volume 4.6 × 10–3 m3. Calculate the new pressure of the gas.

4An ideal gas expands adiabatically from a state with volume 2.8 × 10–3 m3 and temperature 560 K to a state of volume 4.8 × 10–3 m3. Calculate the new temperature of the gas.

5A gas expands at a constant pressure of 5.4 × 105 Pa from a volume of 3.6 × 10–3 m3 to a volume of 4.3 × 10–3 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 pressure–volume 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.