Energy and Entropy: Fall-2024 HW 4 (SOLUTION): Due Day 13 Fri 10/11
Bottle in a Bottle Part 2
S1 5075S
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. The outer bottle is insulated.
The volume of the small bottle is 0.001 m3 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 are initially in the small bottle? What is the final temperature of the gas after the pressures have equalized?
From the equation of state, one gets
\begin{equation*}
N=\frac{p V}{k_{B} T}=\frac{(106 Pa)(0.001 m^3)}{(1.38*10^{-23}\frac{J}{K})(300K)}=2.56 \times 10^{19}\text{ atoms}
\end{equation*}
The final temperature of the gas is 300 K. This free expansion does no work (since there is nothing moving to push against). Furthermore, since since this is a free expansion that does no work, and since outer bottle is insulated, no energy is transfered by heating. This means that the internal energy must not change, and thus the temperature does not change, because \(U=NkT\).
Compute the change of entropy \(\Delta S\) between the initial state (gas in the small bottle) and the final state (gas in both bottles, pressures equalized). Do not use the Sackur-Tetrode equation, use an alternative method.
The big bottle is thermally insulated from its surroundings, and there are no temperature differences inside the big bottle, therefore, \(\int \frac{{\mathit{\unicode{273}}} Q}{T} = 0\) for the leaky-bottle process. However, a free-expansion process, like this one, is not quasi-static. We can't trust \(\int \frac{{\mathit{\unicode{273}}} Q}{T}\) as a measure of entropy change. To figure out \(\Delta S\), we must think of a quasi-static process that goes from the inital state to the final state.
We can pick any quasi-static process that has the same initial and final state. The easiest choice for our calcuation is a slow isothermal expansion. For this alternative process, imagine that the small bottle does not leak. Instead, the small bottle gradually grows to the same size as the large bottle. Heat is allowed to flow from the outside world into the gas, so the gas stays at the same temperature throughout the expansion. We'll keep track of the heat entering the gas during this quasi-static process, and then compute \(\int \frac{{\mathit{\unicode{273}}} Q}{T}\).
\begin{equation*}
T dS=p dV
\end{equation*}
\begin{equation*}
dS=\frac{p}{T}dV=\frac{N k_B}{V} dV
\end{equation*}
Where we used the equation of state in the last step. Integrating we have:
\begin{align*}
\Delta S&=\int_{V_i}^{V_f} \frac{N k_{B}}{V} d V
\\
&=N k_{B} \ln \left(\frac{V_f}{V_i}\right)
\\
&= N k_B\ln(10)
\\
&\approx 8.34*10^{-4} \frac{J}{K}
\end{align*}
You could choose a more tricky path, such as an adiabatic expansion to the final volume followed by a constant-volume heating to the initial temperature. But that would just be a pain.
Discuss your results.
Gas leaking out of the pressurized small bottle to fill the large bottle is a natural, spontaneous, irreversible process. According to the second law of thermodynamics, such a process must be accompanied by an increase of entropy. However, the simple integration of \({\mathit{\unicode{273}}} Q/T\) cannot be applied directly to this process, because the process is not quasi-static.
It's very easy to forget to check for quasi-static processes, since we most often work with quasistatic processes in our homework!
Melting ice lab questions
S1 5075S
These questions relate to the in-class activity where you put ice (prepared at zero degree celcius) into some warm water. You insulated the ice and water inside nested polystyrene cups and closed the lid. You waited a few minutes for the system to reach equilibrium. At equilibruium, all the melted ice and water were at the same temperature. During class, you started to calculate a prediction for this final temperature.
Prediction What is your prediction for the final temperature? List the measured quantities that you used in your calcultion. Show your work.
Measurement What final temperature did you measure? Comment on the magnitude and sources of errors in your experiment and in your prediction.
Change in entropy Calculate the total change in entropy that occured during the overall process. Initial state: warm water and ice. Final state: cool water at a uniform temperature.
If warm water cools down from 313 K to 283 K, in a quasi-static process, the entropy change is
\begin{align}
\Delta S_{water} &= \int \frac{{\mathit{\unicode{273}}} Q}{T}\\
&= \int_{T_i}^{T_f} \frac{C_pdT}{T}\\
&= C_p \ln\left(\frac{T_f}{T_i}\right)\\
&= C_p \ln\left(\frac{283\text{ K}}{313\text{ K}}\right)\\
&= -0.1008C_p\\
& \Delta S_{\text{water}} \approx-117 \frac{J}{K}
\end{align}
where \(C_p = M_{\text{water}}c_p\) is the heat capacity, where \(c_p=4.183 \frac{J}{gK}\), the specific heat of water. The actual water had this initial temperature and final temperature, so the actual change in entropy must be the same as the quasi-static process that I just calculated.
Now, I'll imagine melting the ice in a series of two quasi-static processes: (1) the phase change, (2) warming the water that was formerly ice. For the phase change, imagine keeping temperature at 273 K throughout the melt process. This gives a change in entropy
\begin{align}
\Delta S_{\text{ice,step1}} &= \int \frac{{\mathit{\unicode{273}}} Q}{T}\\
&= \frac{l_f M_{\text{melted}}}{T}\\
&= \frac{(333 \frac{J}{g})(103g)}{273\text{ K}}\\
&= \Delta S_{\text{ice}} \approx 126 \frac{J}{K}
\end{align}
Now, warming up the melted ice from 273 K to 283 K, in a quasi-static process, the entropy change would be
\begin{align}
\Delta S_{\text{ice,step2}} &= \int \frac{{\mathit{\unicode{273}}} Q}{T}\\
&= \int_{T_i}^{T_f} \frac{C_pdT}{T}\\
&= C_p \ln\left(\frac{T_f}{T_i}\right)\\
&= C_p \ln\left(\frac{313\text{K}}{273\text{K}}\right)\\
&= +0.1367C_p\\
& \Delta S_{\text{ice,step2}} \approx+117 \frac{J}{K}
\end{align}
Now, we can add up all the entropies:
\begin{align}
\Delta S_{net}&=\Delta S_{\text{water}}+\Delta S_{\text{ice,step1}} +\Delta S_{\text{ice,step2}}
\\
&=(126 \frac{J}{K})+(-117 \frac{J}{K})+ something
\\
&=9 \frac{J}{K}
\end{align}
So the entropy of our system increased! This is good, because our process was not quasistatic or reversible,so entropy shouldn't be 0 for the process. The 2nd Law tells us it can only go up for an isolated system that has done an irreversible process.
Adiabatic Ideal Gas
S1 5075S
Consider the adiabatic expansion of a simple ideal gas (adiabatic means that no energy is transfered by heating). The internal energy is given by
\begin{align}
U &= C_v T
\end{align}
where you may take \(C_v\) to be a constant---although for a
polyatomic gas such as oxygen or nitrogen, it is
temperature-dependent. The ideal gas law
\begin{align}
pV &= Nk_BT
\end{align}
determines the relationship between \(p\), \(V\) and \(T\). You may take
the number of molecules \(N\) to be constant.
Use the first law to relate the inexact differential for work to the exact differential \(dT\) for an adiabatic process.
Since we're interested in an adiabatic compression, we know
that \(Q\) is zero. So
\begin{align}
dU &= {\mathit{\unicode{273}}} Q + {\mathit{\unicode{273}}} W \\
dU &= {\mathit{\unicode{273}}} W \quad\textit{since it's adiabatic} \\
C_vdT &= {\mathit{\unicode{273}}} W\quad\textit{from the formula for $U$} \\
{\mathit{\unicode{273}}} W &= C_vdT
\end{align}
Find the total differential \(dT\) where \(T\) is a function \(T(p,V)\).
This is just math using the ideal gas law...
\begin{align}
T &= \frac{pV}{Nk_B} \\
dT &= \frac{pdV + Vdp}{Nk_B}
\end{align}
In the previous two sections, we found two formulas involving
\(dT\). Use the additional definition of work \({\mathit{\unicode{273}}} W = -pdV\) to
solve for the relationship between \(p\), \(dp\), \(V\) and \(dV\) for an
adiabatic process.
Our three equations are
\begin{align}
{\mathit{\unicode{273}}} W &= C_vdT \\
dT &= \frac{pdV + Vdp}{Nk_B}\\
{\mathit{\unicode{273}}} W &= -pdV
\end{align}
We can use the first and second to find that
\begin{align}
{\mathit{\unicode{273}}} W &= \frac{C_v}{Nk_B} \left(pdV + Vdp\right)
\end{align}
which we can combine with the third to obtain
\begin{align}
pdV &= - \frac{C_v}{Nk_B} \left(pdV + Vdp\right) \\
pdV + \frac{C_v}{Nk_B}pdV &= - \frac{C_v}{Nk_B} Vdp \\
\left(\frac{C_v}{Nk_B}+1\right) pdV &= - \frac{C_v}{Nk_B} Vdp \\
\left(\frac{C_v}{Nk_B}+1\right) \frac{dV}{V} &= - \frac{C_v}{Nk_B} \frac{dp}{p} \\
\left(1+\frac{Nk_B}{C_v}\right) \frac{dV}{V} &= -\frac{dp}{p}
\end{align}
Integrate the above differential equation to find a
relationship between the initial and final pressure and volume for
an adiabatic process.
At this point, we can integrate both sides of the equation
\begin{align}
\int_{V_i}^{V_f} \left(1+\frac{Nk_B}{C_v}\right) \frac{dV}{V} &= -
\int_{p_i}^{p_f}\frac{dp}{p} \\
\left(1+\frac{Nk_B}{C_v}\right)\left(\ln V_f - \ln V_i\right) &=
\ln p_i - \ln p_f \\
\left(1+\frac{Nk_B}{C_v}\right)\ln\left(\frac{V_f}{V_i}\right)
&=
\ln\left(\frac{p_i}{p_f}\right) \\
\ln\left(\frac{V_f}{V_i}\right)^{1+\frac{Nk_B}{C_v}}
&=
\ln\left(\frac{p_i}{p_f}\right) \\
\left(\frac{V_f}{V_i}\right)^{1+\frac{Nk_B}{C_v}}
&=
\frac{p_i}{p_f} \\
p_f V_f^{1+\frac{Nk_B}{C_v}} &= p_i V_i^{1+\frac{Nk_B}{C_v}}
\end{align}
The quantity \(1+\frac{Nk_B}{C_v}\) in this equation is commonly
referred to as \(\gamma\).
Adiabatic Compression
S1 5075S
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: Looking up the formula in a textbook is not considered a solution at this level. Use only the equations given, fundamental laws of physics, and results you might have already derived from the same set of equations in other homework questions.
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)?
We are considering an adiabatic compression process (no heat
exchange). Unfortunately, we do not know how the pressure behaves
in an adiabatic process. So let us ask what we know about changes
during a small adiabatic compression.
\begin{align}
dU &= {\mathit{\unicode{273}}} W \\
&= -pdV
\end{align}
where I have used the fact that \({\mathit{\unicode{273}}} Q = 0\) for an adiabatic
process. Now let's use the fact that \(U = U(T)\), which gives us
another expression for a small change in energy:
\begin{align}
dU &= \left(\frac{\partial {U}}{\partial {T}}\right)_{V} dT \\
&= C_V dT
\end{align}
where we know that the derivative involved is \(C_V\) (not \(C_p\))
by considering the heating of a fixed-volume system.
Putting these together:
\begin{align}
dU &= dU \\
-pdV &= C_VdT\label{eq:silly}
\end{align}
Now we have to deal with the pesky pressure. Taking the total
differential of the ideal gas law, we find that
\begin{align}
pdV + Vdp &= Nk_B dT\label{eq:silly2}
\end{align}
So now we can eliminate \(dT\) or \(dV\) from our two equations. I'll
eliminate \(dT\) since we've still got a \(p\) hanging around, so I'll solve for \(dT\) in \eqref{eq:silly} and plug into \eqref{eq:silly2}.
\begin{align}
-pdV &= C_V\frac{pdV + Vdp}{Nk_B} \\
-\left(1 + \frac{C_V}{Nk_B}\right) pdV &= \frac{C_V}{Nk_B} Vdp
\\
-\left(1 + \frac{C_V}{Nk_B}\right) \frac{dV}{V} &= \frac{C_V}{Nk_B} \frac{dp}{p}
\\
-\frac{Nk_B + C_V}{C_V} \frac{dV}{V} &= \frac{dp}{p}
\end{align}
Now we integrate both sides... (one of the three things we do with
differentials)
\begin{align}
-\frac{Nk_B + C_V}{C_V} \int_{V_i}^{V_f}\frac{dV}{V} &=
\int_{p_i}^{p_f}\frac{dp}{p}
\\
-\frac{Nk_B + C_V}{C_V} \ln\frac{V_f}{V_i} &= \ln\frac{p_f}{p_i}
\\
\ln\frac{V_f^{-\frac{Nk_B + C_V}{C_V} }}{V_i^{-\frac{Nk_B + C_V}{C_V} }} &= \ln\frac{p_f}{p_i}
\\
\frac{V_f^{-\frac{Nk_B + C_V}{C_V} }}{V_i^{-\frac{Nk_B + C_V}{C_V} }} &= \frac{p_f}{p_i}
\\
p_iV_i^{\frac{Nk_B + C_V}{C_V} } &= p_fV_f^{\frac{Nk_B + C_V}{C_V} }
\\
p_iV_i^{\gamma } &= p_fV_f^{\gamma }
\end{align}
We can now make use of the ideal gas law to see that
\begin{align}
T_iV_i^{\gamma - 1 } &= T_fV_f^{\gamma - 1 }
\end{align}
Thus since \(V_i/V_f = 15\), we can see that
\begin{align}
T_f &= T_i \left(\frac{V_i}{V_f}\right)^{\gamma - 1 }
\\
T_f &= T_i 15^{0.4}
\\
T_f &= 15^{0.4} \cdot 293 K
\end{align}
By what factor does the pressure increase (before fuel injection)?
Solving for the pressure just involves putting this temperature into the ideal gas law with the final pressure.
