Heat and the First Law of Thermodynamics
First law of Thermodynamics
So far we have talked about total system energy consisting of kinetic
energy, potential energy and thermal energy. Keep in mind thermal
energy represents the kinetic and potential energy of the individual
molecules that make up the system whereas you could imagine giving a
block of wood collective potential energy and kinetic energy by moving
the wood around. The thermal energy is not made up of that collective
energy but considers the vibrations of the individual molecules. Since
we are primarily considering ideal gases, we are not too concerned about
this energy (that is, if we have ideal gas in a jar we are not too concerned
about moving the jar around the room and that type of motion will
typically not affect the temperature of the molecules). In what follows
we only consider the thermal energy of the system.
The only way to change the thermal energy of a previously isolated
system is to transfer energy externally through work (symbol Wtot) or
heat (symbol Q). This leads us to the expression of the first law of
thermodynamics (which is just a statement of conservation of energy):
∆𝐸𝑡ℎ = 𝑄 + 𝑊𝑡𝑜𝑡
Work represents a “force through a distance,” whereas you can imagine
heating up an object by putting it on a stove, for example. Both heat and
work are energy transfers, that is, they are not new forms of energy
(although they have units of Joules) but amounts of energy transferred
from one system to another (they change the amount of thermal energy
in an ideal gas). Work and heat can be positive (energy gain) or
negative (energy loss).
Work Done “on” the Gas (Wtot)
Consider compressing an ideal gas inside a cylinder with a piston as
illustrated below.
If we push the piston down (compress the gas), we do what is called
“work done on the gas” Wtot, which is a positive quantity of energy
gained by the ideal gas. If we let go of the piston, the ideal gas pushes
up on the piston and the work done on the gas is now negative. The
ideal gas is now losing energy. Summarizing, for a compression Wtot>0,
and for an expansion Wtot<0. Notice volume MUST change in order to
have non-zero amount of work done on the gas.
Work Done “by” the Gas (W)
For the expansion above, the ideal gas is said to do work on its
surroundings, W. That is called “work done by the gas” and is positive
for an expansion. The work done by the gas is consequently negative
for a compression. There is a simple relation between work done “on”
the gas vs. work done “by” the gas: Wtot=-W. Summarizing, for a
compression W<0, and for an expansion W>0. W will be important
when we consider heat engines as it represents the work output of an
engine.
Calculating Work
Imagine pushing the piston down by a differential amount of
displacement dy. We expect the work done on the gas will be positive
(compression). If the compression occurs with a roughly constant
amount of force, we can find the differential amount of work done
through the dot product:
⃑
𝑑𝑊𝑡𝑜𝑡 = ⃑𝑭 ∙ 𝒅𝒓
Where force is downward and equal to the pressure times the crosssectional area of the piston: ⃑𝑭 = −𝑝𝐴𝒋̂ and the displacement of the gas
⃑ = 𝑑𝑦𝒋̂. Notice there is no negative sign
molecules is also downward 𝒅𝒓
on the dy because dy is a displacement component and is negative
already (dy<0 downward displacement). The dot product gives
𝑑𝑊𝑡𝑜𝑡 = −𝑝𝐴𝑑𝑦𝒋̂ ∙ 𝒋̂ = −𝑝𝐴𝑑𝑦
From the geometry of the small amount of volume displaced
The volume of that tiny cylinder is just dV=Ady so dWtot=-pdV. Notice
this equation gives positive work for a compression because dV<0
(volume is getting smaller). It also works for an expansion, because
dV>0 and dWtot<0 (negative for an expansion). Integrating the
differentials on both sides we get the work done on the gas for a finite
volume change (compression or expansion):
𝑉𝑓
𝑊𝑡𝑜𝑡 = − ∫ 𝑝𝑑𝑉
𝑉𝑖
Notice that the limits of integration are the initial and final volumes and
be careful to include the negative sign. The integral above can be
represented on a pV diagram as the area under curve:
But you have to be careful of the sign of the work. The work for the last
pV diagram is negative because it is an expansion.
Area Method
To find work done on the gas Wtot using the area method you have to
separate the sign from calculating the area:
𝑊𝑡𝑜𝑡 = ±𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒
You choose a + sign for a compression and a – sign for expansion. Then
you calculate the positive area under the curve.
Example: find the Wtot for the process on the pV diagram below
using both the integral and area method and show that they are
equivalent.
One way to handle this problem is to add the works for process #1
and #2, Wtot=Wtot1+Wtot2. Using the integral method we get
𝑉𝑓
𝑉𝑓
𝑊𝑡𝑜𝑡 = −𝑝𝑖 ∫ 𝑑𝑉 − ∫ 𝑝𝑑𝑉 = −𝑝𝑖 (𝑉𝑓 − 𝑉𝑖 )
𝑉𝑖
𝑉𝑓
Notice that Wtot2=0 because the limits of integration are the same
(that is, work for an isochoric process is exactly zero). We could
also pull the pressure out of the integral because it is constant
(isobaric process). Now let’s calculate Wtot2 through the area
method. Wtot=0 because there is no area under the curve for an
isochoric process. The area method gives Wtot1=+AUC=+lw where
the positive length l=Vi-Vf (notice the intial volume is bigger than
the final volume) and the width w=pi. Notice that Wtot=-pi(Vf-Vi),
the same as the integral result.
Cyclic Process
A cycle on a pV diagram looks like a loop, like the triangular loop
illustrated below.
We would like to consider the work done on the gas per cycle (finishing
at the same thermodynamic state as starting). Notice this work can be
split up Wtot=Wtot1+Wtot2+Wtot3. Notice that Wtot2=0 (isochoric) and that
Wtot1>0 (compression) and Wtot3<0 (expansion). Notice that there is
more work (area under the curve) for expansion in this case than
compression. Thus, we expect the total work per cycle to be negative.
Using the area method we get
𝑊𝑡𝑜𝑡1 = +𝑙𝑤 = +𝑝1 (𝑉2 − 𝑉1 )
1
1
2
2
𝑊𝑡𝑜𝑡3 = − (𝑙𝑤 + 𝑏ℎ) = −(𝑝1 (𝑉2 − 𝑉1 ) + (𝑉2 − 𝑉1 )(𝑝2 − 𝑝1 ))
Notice when the two works are added together, the area of the rectangle
drops out and we are left with negative the area of the triangle:
1
𝑊𝑡𝑜𝑡 = − (𝑉2 − 𝑉1 )(𝑝2 − 𝑝1 )
2
In general, for any cycle, the work per cycle is just the area of the shape:
𝑊𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒,𝑡𝑜𝑡 = ±𝐴𝑟𝑒𝑎 𝑜𝑓𝑠ℎ𝑎𝑝𝑒
Where + is for a counter-clockwise cycle and – is for a clockwise cycle.
Cycles will be important in our discussion of heat engines and heat
pumps.
Isothermal Work
Solving for pressure in the ideal gas law gives p=nRT/V. If we assume
an isothermal process then nRT is a constant. We can then substitute
this function into our integral equation for work:
𝑉𝑓
𝑊𝑡𝑜𝑡 = − ∫
𝑉𝑖
𝑉𝑓
𝑛𝑅𝑇
𝑑𝑉
𝑑𝑉 = −𝑛𝑅𝑇 ∫
= −𝑛𝑅𝑇(ln 𝑉𝑓 − ln 𝑉𝑖 )
𝑉
𝑉
𝑉𝑖
Where we have pulled out the constant from the integrand. Using the
identities lnA-lnB=ln(A/B) and –ln(A/B)=ln(B/A) we have
𝑊𝑡𝑜𝑡 = 𝑛𝑅𝑇 ln
𝑉𝑖
𝑉𝑓
Notice that this equation is good for a compression, because the fraction
of volumes Vi/Vf is greater than one, so Wtot is positive (as expected).
For an expansion, the fraction of volumes is less than one, so Wtot is
negative (as expected).
Isochoric Work
As discussed previously, there is no area under an isochord so Wtot=0.
Isobaric Work
The work done on the gas for an isobaric process can be found using the
area method (it is the area of a rectangle on a pV diagram). For a general
isobaric process, Wtot=-pΔV.
Heat
When heat energy is transferred to a system, its temperature usually
increases in direct proportion to the amount of heat Q added: Q=CΔT,
where C is called the heat capacity. More often we deal with the
specific heat c of a system defined by Q=mcΔT, where m is the mass.
This statement essentially says that if you want to increase the
temperature by a certain about, the heat energy required to do this will
be greater the larger the mass of the object or the larger the specific heat.
Specific heat is a property of a particular material. For example, the
specific heat of metal (a good heat conductor) is much smaller than an
equal mass of Styrofoam (a poor heat conductor). That means it takes a
lot more Q to change the temperature of the Styrofoam.
Units of Specific Heat
From the specific heat equation, its units are J/kgK. The units can also
be listed as J/kg°C because remember that the conversion factor from
Kelvin temperature change to Celcius temperature change is just unity.
Calorimetry
A useful way to determine the specific heat of a substance is through
calorimetry. A simple calorimeter is just a cup of water. Suppose we
know the mass of the water mw and its temperature Tw. We can also look
up the specific heat of liquid water (cw=4190 J/kgK). Suppose also we
have a block of mystery material of mass mx and temperature Tx where
the material is very hot (that is, Tx>>Tw). We would like to find the
specific heat of the metal cx. If we put the material in the water and wait
until equilibrium at temperature Tf we notice that heat energy flows from
hot to cold. That is Qcold is the positive heat gained by the water and Qhot
is the negative heat lost by the material. By conservation of energy we
expect Qhot=-Qcold, that is the energy lost by the material is gained by the
water. Notice the negative sign ensures that the heats have the
appropriate signs for loss and gain. Substituting the heat equations:
𝑚𝑥 𝑐𝑥 ∆𝑇𝑥 = −𝑚𝑤 𝑐𝑤 ∆𝑇𝑤
Substituting the temperatures:
𝑚𝑥 𝑐𝑥 (𝑇𝑓 − 𝑇𝑥 ) = −𝑚𝑤 𝑐𝑤 (𝑇𝑓 − 𝑇𝑤 )
And solve for cx:
𝑐𝑥 =
−𝑚𝑤 𝑐𝑤 (𝑇𝑓 − 𝑇𝑤 )
𝑚𝑥 (𝑇𝑓 − 𝑇𝑥 )
Notice that cx is positive (always for specific heat). Sometimes the
calorimeter will include the cup, as the cup gains heat as well. You will
have to add the contribution to the cup for Qcold knowing what the
specific heat of the cup is.
Latent Heat
The heat associated with a phase (or state) change of a system is called
latent heat. This is the energy to melt or freeze a substance or to boil or
condense a substance. It occurs at constant temperature:
𝑄 = ±𝑚𝐿
Where L is called the latent heat (units of L are J/kg) and the + sign is
for melting and/or boiling and the – sign is for freezing and/or
condensing. Notice if you have water ice at 0°C and you apply the
latent heat of fusion you will have liquid water at 0°C (no temperature
change).
Example: what is the heat energy Q required to change 1g of water
ice at Ti=-30°C to steam at Tf=120°C?
Part A: heat the ice to freezing temperature 0°C:
QA=miciΔTi=(0.001kg)(2090J/kgK)(0°C+30°C)=62.7J
Part B: apply latent heat of fusion to melt the ice at 0°C:
QB=+miLf=(0.001kg)(3.33x105J/kg)=333J
Part C: heat the liquid water to 100°C:
QC=micwΔTw=(0.001kg)(4190J/kgK)(100°C-0°C)=419J
Part D: apply the latent heat of vaporization to boil the water to
steam at 100°C:
QD=+miLv=(0.001kg)(2.26x106J/kg)=2260J
Part E: heat the steam to 120°C:
QE=micsΔTs=(0.001kg)(2.01x103J/kgK)(120°C-100°C)=40.2J
The total heat is a sum of all the heats above Q=3114.9J. We can
make a plot of the heat added as a function of temperature:
Notice that changing the liquid water to steam requires the most
heat energy. Steam is very dangerous because it holds a lot of
energy!
Molar Specific Heat and Thermal Energy
Consider the two isotherms on the pV diagram below.
Notice that T2>T1 and that there are an infinite number of ways to jump
from the cold isotherm to the hot isotherm. Two ways are shown above.
Path A is an isochoric (constant volume) process and path B is an
isobaric (constant pressure) process. From the last chapter, the only way
to change the thermal energy of an ideal gas is to change the
temperature. Thus, the change in thermal energy for the two paths are
equal: ΔEthA=ΔEthB because the temperature change is the same. From
the first law, we can find these changes in thermal energy.
We need to find isobaric and isochoric heat first and they are given by
molar specific heat:
𝑄 = 𝑛𝐶𝑉 ∆𝑇 𝑎𝑛𝑑 𝑄 = 𝑛𝐶𝑝 ∆𝑇
These equations are only good for isochoric or isobaric processes. The
capital C’s in the equation are the molar specific heats for constant
volume and pressure. Their units are J/molK. Values will be discussed
soon.
Now we can find the thermal energies for each path and set them equal
to each other. For the isochoric process (A) the first law requires just
heat (work is zero):
∆𝐸𝑡ℎ𝐴 = 𝑄 = 𝑛𝐶𝑉 ∆𝑇
For the isobaric process (B) we have
∆𝐸𝑡ℎ𝐵 = 𝑄 + 𝑊𝑡𝑜𝑡 = 𝑛𝐶𝑝 ∆𝑇 − 𝑝∆𝑉
Setting the two equal we get
𝑛𝐶𝑉 ∆𝑇 = 𝑛𝐶𝑝 ∆𝑇 − 𝑝∆𝑉
From the ideal gas law for an isobaric process we can write changes like
this:
𝑝∆𝑉 = 𝑛𝑅∆𝑇
And substituting it into our thermal energy equation we get
𝑛𝐶𝑉 ∆𝑇 = 𝑛𝐶𝑝 ∆𝑇 − 𝑛𝑅∆𝑇
Dividing out the nΔT’s we arrive at a very simple relation for the molar
specific heats: Cp=CV+R.
Another interesting relation deals with the change in thermal energy.
Remember that any path from cold to hot isotherm gives the same
change in thermal energy so for any ideal gas process:
∆𝐸𝑡ℎ = 𝑛𝐶𝑉 ∆𝑇
Values for Specific Heats
For a monatomic ideal gas Cv=(3/2)R and Cp=(5/2)R. For a diatomic
ideal gas at room temperature CV=(5/2)R and Cp=(7/2)R. We will see
where these come from soon.
Shortcut for getting isobaric and isochoric heats
The pV diagram below shows an isochoric process followed by an
isobaric process. Suppose also that the gas is monatomic:
To find the heats for these processes (1 and 2), you might need to know
the number of moles and the temperatures. However, there is a shortcut
to finding the heats for these processes. For example, for the isochoric
process we have Q1=nCVΔT. From changes in the ideal gas law for an
isochoric process we have ΔpV=nRΔT or nΔT=ΔpV/R so Q1=CvΔpV/R.
Using, for a monatomic gas, CV=(3/2)R we get Q1=(3/2)ΔpV. Likewise,
using a similar approach for the isobaric process Q2=(5/2)ΔVp.
Summarizing, to get the heat you find the fraction you need, multiply by
Δ the thing that changes times the thing that remains constant.
Notice if this gas was diatomic instead, the heats would be given by
Q1=(5/2)ΔpV and Q2=(7/2)ΔVp. That is, the fraction would change.
Notice also that you can tell the signs of the heats. Because the pressure
is decreasing Q1 is negative (loss) and because the volume is increasing
Q2 is positive (gain).
Example: suppose we have a monatomic ideal gas using the
previous pV diagram where p2=10Pa, p1=6Pa, V1=5m3, and
V2=10m3. Find Q1, Wtot1, Q2, and Wtot2.
Q1=(3/2)Δp1V1=(3/2)(6Pa-10Pa)(5m3)=-30J
Wtot1=0 (isochoric)
Q2=(5/2) ΔV2p1=(5/2)(10m3-5m3)(6Pa)=75J
Wtot2=-lw=-(10m3-5m3)(6Pa)=-30J
Isothermal Heat
The change in thermal energy is zero for an isothermal process. The
first law gives
∆𝐸𝑡ℎ = 𝑊𝑡𝑜𝑡 + 𝑄 = 0
This does not mean there is no heat and work, it just means that Wtot=-Q,
that is, any heat that goes into a system leaves by work and vice versa
for an isothermal process. From what we know about work,
Q=nRTln(Vf/Vi) for an isothermal process.
Adiabatic Process
An adiabatic process is when Q=0. One can imagine this process
occurring in an insulated container where no heat can escape. An
adiabatic looks steeper than a corresponding isothermal process on a pV
diagram because of the equations connecting the initial and final
thermodynamic states:
𝛾
𝛾
𝑝𝑖 𝑉𝑖 = 𝑝𝑓 𝑉𝑓
𝛾−1
𝑇𝑖 𝑉𝑖
𝛾−1
= 𝑇𝑓 𝑉𝑓
Where γ is the adiabatic index and is obtained through the molar
specific heats γ=Cp/CV. For a monatomic gas γ=5/3 and for a diatomic
gas γ=7/5. It’s the adiabatic index (the exponent) that makes the curve
steeper than a corresponding isotherm.
From the first law of thermodynamics ΔEth=Wtot for an adiabatic
process. Since all ideal gas processes have the same change in thermal
energy, Wtot=nCvΔT for an adiabatic process only.
Adiabatic Free Expansion
A very special type of adiabatic process is the adiabatic free expansion.
Imagine two insulated chambers connected by a thin membrane as
illustrated below. The process is adiabatic because no heat can flow
(insulated). In one chamber there is a vacuum and the other ideal gas.
Imagine poking a hole in the membrane. The molecules are moving and
eventually the gas will fill the entire chamber (the “AFTER” case).
From the first law we have ΔEth=0, that is, there is no heat OR work.
There is no work because the gas molecules don’t push and move
anything up (like a piston).
Since there is no change in thermal energy, there is no change in
temperature, so this expansion is not only adiabatic but isothermal also!
Summarizing, for the adiabatic free expanstion, ΔEth=0, ΔT=0, Wtot=0,
and Q=0. Notice also that by p=nRT/V, since nRT is constant, if the
volume doubles the pressure decreases by half.
Another way to think about the fact that this is an isothermal process is
that poking a hole in the membrane does not change the rms speed of the
molecules. We know that temperature is only a function of rms speed,
so if the speed does not change the temperature does not change.
Degrees of Freedom and Equipartition
Recall that the average kinetic energy per molecule was only a function
of the rms speed of the molecules. The expression below is only true for
a monatomic gas (or a gas that behaves like a monatomic gas):
1
3
2
𝑚𝑣𝑟𝑚𝑠
= 𝑘𝐵 𝑇
2
2
We can write out the rms speed in terms of its components:
1
3
2
2
2
𝑚(𝑣𝑟𝑚𝑠,𝑥
+ 𝑣𝑟𝑚𝑠,𝑦
+ 𝑣𝑟𝑚𝑠,𝑧
) = 𝑘𝐵 𝑇
2
2
Notice in a random distribution all the rms components are equal and
requires that each one contributes exactly 1/2kBT to the kinetic energy
per molecule. This is called equipartition of energy, that is, each degree
of freedom contributes exactly 1/2kBT to the average kinetic energy per
molecule. The symbol for degree of freedom is f. A degree of freedom
is a “way of moving”. Notice there are f=3 degrees of freedom for a
monatomic gas. That is, the individual gas molecule (which is an atom)
can move in 3 ways (x, y and z direction).
More complicated gases (like diatomic gases) have more ways of
moving (at certain temperatures) and we have to amend our energy
equations to account for extra degrees of freedom:
𝑓
𝑘 𝑇
2 𝐵
𝑓
= 𝑁𝐾𝑎𝑣𝑔 = 𝑁𝑘𝐵 𝑇
2
𝐾𝑎𝑣𝑔 =
𝐸𝑡ℎ
Diatomic Gas
A diatomic gas molecule (like diatomic nitrogen) consists of two gas
molecules connected by a chemical bond. The bond can produce extra
degrees of freedom because at higher temperatures, motion about the
bond can be created (rotations and vibrations). Typically, for a very
cold diatomic gas (10’s of Kelvin) f=3, and it acts like a monatomic gas.
For temperatures closer to room temperature (around 300K) f=5, the
molecule can start to rotate about two axes. For even higher
temperatures (around 1000K) the molecules can start to vibrate in two
ways (slow and fast) and f=7.
Molar Specific Heats and Adiabatic Index
From our amended thermal energy equation we can convert to the molar
version of changes in thermal energy:
∆𝐸𝑡ℎ =
𝑓
𝑓
𝑁𝑘𝐵 ∆𝑇 = 𝑛𝑅∆𝑇 = 𝑛𝐶𝑉 ∆𝑇
2
2
The last equation remember we found was good for ANY ideal gas
process. Comparing the two equations to the right, we see that
CV=(f/2)R and from that we can get Cp, that is, Cp=CV+R. Thus, for a
monatomic gas we get CV=(3/2)R and Cp=(5/2)R. Likewise, for a
diatomic gas we get f=5 so CV=(5/2)R and Cp=(7/2)R. The molar
specific heats also give us the adiabatic index γ=Cp/CV. So γ=5/3 for
monatomic and γ=7/5 for diatomic (at room temperature).
Conduction of Heat
We’ve seen that heat energy moves from hot to cold and, when you heat
something up, it results in a temperature change. This movement is
called “conduction”. Conduction is described by the following power
formula:
𝑃=
𝑄
𝐴
= 𝑘 (𝑇ℎ𝑜𝑡 − 𝑇𝑐𝑜𝑙𝑑 )
∆𝑡
𝑙
Where Q is the heat that flows from one system to the other, A is the
cross sectional area of the “path” that the heat flows, l is the distance the
heat travels, k is called the “thermal conductivity,” and Δt is the time it
takes for the heat to flow. Notice, you get more heat in a smaller
amount of time for a bigger temperature difference, also more heat can
flow if the A is bigger (a bigger channel for the heat to flow). More heat
can flow for a bigger thermal conductivity, which is a property of
various materials. Finally, less heat per unit time flows if the path is
longer as it takes a longer time for the energy to reach the object.
Radiation
Light sources, like a light bulb or the Sun radiate energy away from the
source through the Stefan-Boltzmann law:
𝑃=
𝑄
= 𝜎𝜖𝐴𝑇 4
∆𝑡
Where σ=5.6703x10-8 W/m2K4 is known as the Stephan-Boltzmann
constant and ε is called emissivity which is a number from 0 to 1, if ε is
close to 1 the source is a good emitter AND absorber, that is it can
absorb radiation just as good as it can emit it (a dark object has a high
emissivity, for example).