A diesel
engine requires no spark plug. Rather, the air in the cylinder is
compressed so highly that the fuel ignites spontaneously when
sprayed into the cylinder.
In this problem, you may treat air as an ideal gas, which satisfies
the equation \(pV = Nk_BT\). You may also use the property of an
ideal gas that the internal energy depends only on the temperature
\(T\), i.e. the internal energy does not change for an isothermal
process. For
air at the relevant range of temperatures the heat capacity at
fixed volume is given by \(C_V=\frac52Nk_B\), which means the internal energy
is given by \(U=\frac52Nk_BT\).
Note: in this problem you are expected to use only the equations
given and fundamental physics laws. Looking up the formula in a textbook
is not considered a solution at this level.
If the air is initially at room temperature
(taken as \(20^{o}C\)) and is then compressed adiabatically to \(\frac1{15}\)
of the original volume, what final temperature is attained (before fuel
injection)?
Consider the bottle in a bottle problem in a previous problem set, summarized here.
A small bottle of
helium is placed inside a large bottle, which otherwise contains
vacuum. The inner bottle contains a slow leak, so that the helium
leaks into the outer bottle. The inner bottle contains one tenth
the volume of the outer bottle, which is insulated.
The volume of the small bottle is 0.001 m23 and the volume of the big bottle is 0.01 m3. The initial state of the gas in the small bottle was \(p=106\) Pa and its temperature \(T=300\) K. Approximate the helium gas as an ideal gas of equations of state \(pV=Nk_BT\) and \(U=\frac32 Nk_BT\).
How many molecules of gas does the large bottle contain? What is the final temperature of the gas?
Compute the integral \(\int \frac{{\mathit{\unicode{273}}} Q}{T}\) and the change of entropy \(\Delta S\) between the initial state (gas in the small bottle) and the final state (gas leaked in the big bottle).
Students will determine the change in entropy (positive, negative, or none) for both the system and surroundings in three different cases. This is followed by an active whole-class discussion about where the entropy comes from during an irreversible process.
Plot your data I
Plot the temperature versus total energy
added to the system (which you can call \(Q\)). To do this, you will
need to integrate the power. Discuss this curve and any interesting
features you notice on it.
Plot your data II
Plot the heat capacity versus
temperature. This will be a bit trickier. You can find the heat
capacity from the previous plot by looking at the slope.
\begin{align}
C_p &= \left(\frac{\partial Q}{\partial T}\right)_p
\end{align}
This is what is called the heat capacity, which is the amount
of energy needed to change the temperature by a given amount. The
\(p\) subscript means that your measurement was made at constant
pressure. This heat capacity is actually the total heat capacity of
everything you put in the calorimeter, which includes the resistor
and thermometer.
Specific heat From your plot of \(C_p(T)\), work out the
heat capacity per unit mass of water. You may assume the effect of
the resistor and thermometer are negligible. How does your answer
compare with the prediction of the Dulong-Petit law?
Latent heat of fusion
What did the temperature do while the ice was melting? How much
energy was required to melt the ice in your calorimeter? How much
energy was required per unit mass? per molecule?
Entropy of fusion The change in entropy is easy to measure for a
reversible isothermal process (such as the slow melting of ice),
it is just
\begin{align}
\Delta S &= \frac{Q}{T}
\end{align}
where \(Q\) is the energy thermally added to the system and \(T\) is
the temperature in Kelvin. What is was change in the entropy of
the ice you melted? What was the change in entropy per
molecule? What was the change in entropy per molecule divided
by Boltzmann's constant?
Entropy for a temperature change Choose two temperatures
that your water reached (after the ice melted), and find the change
in the entropy of your water. This change is given by
\begin{align}
\Delta S &= \int \frac{{\mathit{\unicode{273}}} Q}{T} \\
&= \int_{t_i}^{t_f} \frac{P(t)}{T(t)}dt
\end{align}
where \(P(t)\) is the heater power as a function of time and \(T(t)\) is
the temperature, also as a function of time.
Students work out heat and work for rectangular paths on \(pV\) and \(TS\) plots. This gives with computing heat and work, applying the First Law, and recognizing that internal energy is a state function, which cannot change after a cyclic process.
In this introduction to heat capacity, students determine a derivative that indicates how much the internal energy changes as the temperature changes when volume is held constant.
Nuclei of a particular isotope species contained in a crystal have
spin \(I=1\), and thus, \(m = \{+1,0,-1\}\). The interaction between
the nuclear quadrupole moment and the gradient of the crystalline
electric field produces a situation where the nucleus has the same
energy, \(E=\varepsilon\), in the state \(m=+1\) and the state \(m=-1\),
compared with an energy \(E=0\) in the state \(m=0\), i.e. each nucleus
can be in one of 3 states, two of which have energy \(E=\varepsilon\)
and one has energy \(E=0\).
Find the Helmholtz free energy \(F = U-TS\) for a crystal
containing \(N\) nuclei which do not interact with each other.
Find an expression for the entropy as a function of
temperature for this system. (Hint: use results of part a.)
Indicate what your results predict for the entropy at the
extremes of very high temperature and very low temperature.
Suppose that a
system of \(N\) atoms of type \(A\) is placed in diffusive contact
with a system of \(N\) atoms of type \(B\) at the same temperature and
volume.
Show that after diffusive equilibrium is reached the total entropy
is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is
known as the entropy of mixing.
If the atoms are identical (\(A=B\)), show that there is no increase
in entropy when diffusive contact is established. The difference has
been called the Gibbs paradox.
Since the Helmholtz free energy is lower for the mixed \(AB\) than
for the separated \(A\) and \(B\), it should be possible to extract
work from the mixing process. Construct a process that could extract
work as the two gasses are mixed at fixed temperature. You will
probably need to use walls that are permeable to one gas but not the
other.
Note
This course has not yet covered work, but it was covered in
Energy and Entropy, so you may need to stretch your memory to finish
part (c).
Energy and Entropy 2021 (2 years)
The internal energy is of any ideal gas can be written as
\begin{align}
U &= U(T,N)
\end{align}
meaning that the internal energy depends only on the number of
particles and the temperature, but not the volume.* The ideal gas law
\begin{align}
pV &= Nk_BT
\end{align}
defines the relationship between \(p\), \(V\) and \(T\). You may take the
number of molecules \(N\) to be constant. Consider the free adiabatic
expansion of an ideal gas to twice its volume. “Free expansion”
means that no work is done, but also that the process is also
neither quasistatic nor reversible.
What is the change in entropy of the gas? How do you know
this?
