Energy and Entropy: Fall-2024 HW 3 (SOLUTION): Due Day 10 Tues 10/8
Name the Experiment
S1 5073S
Consider the following derivative:
\begin{align}
\left(\frac{\partial {U}}{\partial {p}}\right)_{S,N}
\end{align}
Design an experiment to measure this derivative. In your answer, include a schematic of the aperatus and label the quantitites you would measure. Describe your measurement process, and show the algebra required to convert your directly measured quantities into the derivative.
To hold the entropy fixed we want an insulated piston. To control the pressure it will be convenient to have a stand on top of the piston that we can add or remove weights from. So our experiment will involve adding a bit of mass to the piston, which gives us a change of pressure. We will measure the resulting change in volume, \(\Delta V\), by measuring the change in height of the piston, \(\Delta z\), and the area of the plunger that contacts the gas, \(A\). We still need to figure out how the internal energy changes. Going to the First Law or the thermodynamic identity
\begin{align}
dU &= {\mathit{\unicode{273}}} Q + {\mathit{\unicode{273}}} W \\
&= TdS - pdV
\end{align}
Since there is no heating and no change in entropy, we can see that
\begin{align}
dU &= -pdV
\end{align}
i.e. the change in internal energy is just the work done. So the derivative
is given by
\begin{align}
\left(\frac{\partial {U}}{\partial {p}}\right)_{S} &\approx \frac{-p\Delta V}{\Delta p}
\\
&= \frac{-pA\Delta z}{\Delta F/A}
\end{align}
where \(A\) is the area. I left the total pressure as \(p\) rather than \(F/A\)
because the total pressure isn't just the force due to our weights over the
area, but also includes the pressure from the atmosphere.
Extensive Internal Energy
S1 5073S
Consider a system which has an internal energy \(U\) defined by:
\begin{align}
U &= \gamma V^\alpha S^\beta
\end{align}
where \(\alpha\), \(\beta\) and \(\gamma\) are constants. The internal
energy is an extensive quantity. What constraint does this place on
the values \(\alpha\) and \(\beta\) may have?
Since \(U\), \(V\) and \(S\) are all extensive, if you scale up the
system by a factor \(\lambda\), they each scale up by the same
factor. Thus when we scale everything up by \(\lambda\) (except the
constants, which are constant), we find:
\begin{align}
U(S,V) &= \gamma V^\alpha S^\beta \\
U(\lambda S, \lambda V) &= \gamma(\lambda V)^\alpha(\lambda
S)^\beta \\
\lambda U(S,V) &= \lambda^{\alpha + \beta}\gamma V^\alpha
S^\beta \\
\lambda &= \lambda^{\alpha + \beta} \\
\alpha + \beta &= 1
\end{align}
Rubber Sheet
S1 5073S
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 vertical dimension of the rubber sheet we will call \(y\), and the horizontal dimension of the rubber sheet we will call \(x\). We can use these two independent variables to specify the "state" of the rubber sheet. Similiar to the partial derivative machine, we could choose any pair of variables from the set \(\{ x,y,F_x,F_y \}\) to specify the state of the rubber sheet.
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 \(U(x,y)\).
We are told that \(U=U(x,y)\), i.e. \(U\) is a function of \(x\) and \(y\). Thus, a mathematically true statement is
\begin{equation}
dU = \left(\frac{\partial {U}}{\partial {x}}\right)_{y} dx + \left(\frac{\partial {U}}{\partial {y}}\right)_{x} dy.
\end{equation}
From the physics of work (work = force \(\times\) distance moved parallel to the force), I know that
\begin{equation}
dU = F_x dx + F_y dy.
\end{equation}
Comparing these equations, I conclude that
\begin{align}
F_x &= \left(\frac{\partial {U}}{\partial {x}}\right)_{y}.
\end{align}
I will need this expression for \(F_x\) later. The system has excess variables which gives me multiple options for my choice of 2 independent variables. For example, it is fine to write \(x(y,F_x)\), i.e. think of \(x\) as a function of \(y\) and \(F_x\). Thus, a mathematically true statement is
\begin{equation}
dx = \left(\frac{\partial {x}}{\partial {y}}\right)_{F_x} dy + \left(\frac{\partial {x}}{\partial {F_x}}\right)_{y} dF_x
\end{equation}
In our experiment, we won't let \(F_x\) change, which is the same as saying \(dF_x = 0\). So, relating back to our experimentally measured distances, I can write:
\begin{equation}
\Delta x = \left(\frac{\partial {x}}{\partial {y}}\right)_{F_x} \Delta y
\end{equation}
Now, I'd like to find a way to express \(\left(\frac{\partial {x}}{\partial {y}}\right)_{F_x}\) in terms of partial derivatives of \(U\). My strategy is to write down the other mathematically true statements relating \(dx\), \(dy\) and \(dF_x\):
\begin{equation}
dy = \left(\frac{\partial {y}}{\partial {x}}\right)_{F_x} dx + \left(\frac{\partial {y}}{\partial {F_x}}\right)_{x} dF_x\\
dF_x = \left(\frac{\partial {F_x}}{\partial {x}}\right)_{y} dx + \left(\frac{\partial {F_x}}{\partial {y}}\right)_{x} dy
\end{equation}
The second of these two equations is useful. It can be rearranged to give
\begin{align}
dx &= -\frac{\left(\frac{\partial {F_x}}{\partial {y}}\right)_{x}}{\left(\frac{\partial {F_x}}{\partial {x}}\right)_{y}}dy +
\frac{1}{\left(\frac{\partial {F_x}}{\partial {x}}\right)_{y}}dF_x \\
\end{align}
The first term on the right hand side must be equal to \(\left(\frac{\partial {x}}{\partial {y}}\right)_{F_x}dy\). Thus we find our derivative:
\begin{align}
\left(\frac{\partial {x}}{\partial {y}}\right)_{F_x} &=
-\frac{\left(\frac{\partial {F_x}}{\partial {y}}\right)_{x}}{\left(\frac{\partial {F_x}}{\partial {x}}\right)_{y}}.
\end{align}
Substituting for \(F_x\) we see that
\begin{align}
\Delta x &=
-\frac{\left(\frac{\partial {\left(\frac{\partial {U}}{\partial {x}}\right)_{y}}}{\partial {y}}\right)_{x}}{\left(\frac{\partial {\left(\frac{\partial {U}}{\partial {x}}\right)_{y}}}{\partial {x}}\right)_{y}}
\Delta y
\end{align}
It's OK if you simplify the notation and write
\begin{align}
-\frac{\frac{\partial^2U}{\partial x\partial
y}}{\left(\frac{\partial^2 U}{\partial x^2}\right)_{y}} \Delta y
\end{align}
99.9 % of physicists will understand from this notation what you are holding constant during each of these partial derivatives.
Bottle in a Bottle
S1 5073S
The internal energy of helium gas at temperature \(T\) is
to a very good approximation given by
\begin{align}
U &= \frac32 Nk_BT
\end{align}
Consider a very irreversible process in which 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. What is the
change in temperature when this process is complete? What fraction of the
helium will remain in the small bottle?
There is no work done because nothing is moving. Since the bottle
is insulated, there is no heat either, and thus from the first law
\begin{align}
\Delta U &= Q - W
\end{align}
the internal energy is unchanged. Since \(U = \frac32 N k_B T\), we
conclude that the temperature is unchanged.
The pressure in the two bottles will be the same, so there will be
one tenth of the gas in the small bottle, with the rest in the large
bottle. Except, of course, that the gas that is in the small bottle
is also in the large bottle.
