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).
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)?
Energy and Entropy 2021 (2 years)
We have the following equations of state for the total magnetization \(M\), and the entropy \(S\) of a paramagnetic system:
\begin{align}
M&=N\mu\, \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}}
{e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}\\
S&=Nk_B\left\{\ln 2 +
\ln \left(e^{\frac{\mu B}{k_B T}}+e^{-\frac{\mu B}{k_B T}}\right)
+\frac{\mu B}{k_B T}
\frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}}
{e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}
\right\}
\end{align}
List variables in their proper positions in the middle columns of the charts below.
Solve for the magnetic susceptibility, which is defined as:
\[\chi_B=\left(\frac{\partial M}{\partial B}\right)_T
\]
Using both the differentials (zapping with d) and chain rule diagram methods, find a chain rule for:
\[\left(\frac{\partial M}{\partial B}\right)_S
\]
Evaluate your chain rule. Sense-making: Why does this come out to zero?
In class, you measured the isolength
stretchability and the isoforce stretchability of your systems in the
PDM. We found that for some systems these were very
different, while for others they were identical.
Show with algebra (NOT experiment) that the ratio of isolength stretchability to isoforce
stretchability is the same for both the left-hand side of the system and the right-hand side of the system.
i.e.:
\begin{align}
\frac{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{x_R}}{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{F_R}} &=
\frac{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{x_L}}{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{F_L}}
\label{eq:ratios}
\end{align}
Hint
You will need to make use of the cyclic chain
rule:
\begin{align}
\left(\frac{\partial {A}}{\partial {B}}\right)_{C} = -\left(\frac{\partial {A}}{\partial {C}}\right)_{B}\left(\frac{\partial {C}}{\partial {B}}\right)_{A}
\end{align}
Hint
You will also need the ordinary chain
rule:
\begin{align}
\left(\frac{\partial {A}}{\partial {B}}\right)_{D} = \left(\frac{\partial {A}}{\partial {C}}\right)_{D}\left(\frac{\partial {C}}{\partial {B}}\right)_{D}
\end{align}
Consider a hanging rectangular rubber sheet.
We will consider there to be two ways to get energy into or out of
this sheet: you can either stretch it vertically or horizontally.
The distance of vertical stretch we will call \(y\), and the distance
of horizontal stretch we will call \(x\).
If I pull the bottom down by a small distance \(\Delta y\), with no
horizontal force, what is the resulting change in width \(\Delta x\)?
Express your answer in terms of partial derivatives of the potential
energy \(U(x,y)\).
Static Fields 2023 (5 years)
This short small group activity introduces students to the Leibniz notation used for partial derivatives in thermodynamics; unlike standard Leibniz notation, this notation explicitly specifies constant variables. Students are guided in linking the variables from a contextless Leibniz-notation partial derivative to their proper variable categories.
thermodynamicsMaxwell relationEnergy and Entropy 2020
The Gibbs free energy,
\(G\), is given by
\begin{align*}
G = U + pV - TS.
\end{align*}
Find the total differential of \(G\). As always, show your work.
Interpret the coefficients of the total differential \(dG\) in
order to find a derivative expression for the entropy \(S\).
From the total differential \(dG\), obtain a different
thermodynamic derivative that is equal to
\[ \left(\frac{\partial {S}}{\partial {p}}\right)_{T} \]
In economics, the term utility is roughly related to overall
happiness. Many things affect your happiness, including the amount of
money you have and the amount of coffee you drink. We cannot directly
measure your happiness, but we can measure how much money you
are willing to give up in order to obtain coffee or bagels. If we
assume you choose wisely, we can thus determine that your happiness
increases when you decrease your amount of money by that amount in
exchange for increasing your coffee consumption. Thus money is a
(poor) measure of happiness or utility.
Money is also a nice quantity because it is conserved---just like
energy! You may gain or lose money, but you always do so by a
transaction. (There are some exceptions to the conservation of money,
but they involve either the Fed, counterfeiters, or destruction of
cash money, and we will ignore those issues.)
In this problem, we will assume that you have bought all the coffee
and bagels you want (and no more), so that your happiness has been
maximized. Thus you are in equilibrium with the coffee shop. We will
assume further that you remain in equilibrium with the coffee shop at
all times, and that you can sell coffee and bagels back to the coffee
shop at cost.*
Thus your savings \(S\) can be considered to be a function of your
bagels \(B\) and coffee \(C\). In this problem we will also discuss the
prices \(P_B\) and \(P_C\), which you may not assume are
independent of \(B\) and \(C\).
It may help to imagine that you could possibly buy out the local supply of coffee,
and have to import it at higher costs.
The prices of bagels and coffee \(P_B\) and \(P_C\) have derivative
relationships between your savings and the quantity of coffee and
bagels that you have. What are the units of these prices? What is
the mathematical definition of \(P_C\) and \(P_B\)?
Write down the total differential of your savings, in terms of
\(B\), \(C\), \(P_B\) and \(P_C\).
Solve for the total differential of your net worth. Your net worth
\(W\) is the sum of your total savings plus the value of the coffee and
bagels that you own. From the total differential, relate your amount of
coffee and bagels to partial derivatives of your net worth.
Energy and Entropy 2021 (2 years)
The isothermal
compressibility is defined as
\begin{equation}
K_{T}=-\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{T}
\end{equation}
\(K_T\) is be found by measuring the fractional change in volume when
the the pressure is slightly changed with the temperature held
constant. In contrast, the adiabatic compressibility is defined as
\begin{equation}
K_{S}=-\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{S}
\end{equation}
and is measured by making a slight change in pressure without
allowing for any heat transfer. This is the compressibility, for
instance, that would directly affect the speed of sound. Show that
\begin{equation}
\frac{K_{T}}{K_{S}} = \frac{C_{p}}{C_{V}}
\end{equation}
Where the heat capacities at constant pressure and volume are given
by
\begin{align}
C_{p} &= T \left(\frac{\partial S}{\partial T}\right)_{p} \\
C_{V} &= T \left(\frac{\partial S}{\partial T}\right)_{V}
\end{align}
