Energy and Entropy: Winter-2026 HW 3 (SOLUTION): Due Day 14 W5 D3
Entropy and Temperature
S1 5455S
Suppose the multiplicity of a system is
\(\Omega(U) = CU^{3N/2}\), where \(C\) is a constant and \(N\) is the
number of particles.
Show that \(U=\frac32 N k_BT\).
We begin by finding the entropy given the provided multiplicity.
\begin{align}
S(U,N) &= k_B\log \Omega(U,N) \\
&= k_B \log \left(CU^{3N/2}\right) \\
&= k_B\left( \log C + \log \left(U^{3N/2}\right) \right) \\
&= k_B\log C + \frac32 Nk_B \log U \\
\end{align} In the last two steps there, we made use of properties
of \(\log\). If these are not obvious to you, you absolutelymust take the time to review the properties of logarithms.
They are absolutely critical to this course! \begin{align}
\frac1{T} &= \left(\frac{\partial S}{\partial U}\right)_{V,N}
\\
&= \frac32Nk_B\frac1{U}
\\
U &= \frac32Nk_BT
\end{align} Yay.
Show that \(\left(\frac{\partial^2S}{\partial U^2}\right)_N\) is
negative. This form of \(\Omega(U)\) actually applies to a monatomic
ideal gas.
We just need to take one more derivative, since we already found
\(\left(\frac{\partial S}{\partial U}\right)_{V,N}\) in part (a).
\begin{align}
\left(\frac{\partial^2S}{\partial U^2}\right)_{V,N} &=
-\frac32Nk_B\frac1{U^2} \\
&< 0,
\end{align} where in the last step we only needed to assume that
\(N>0\) (natural for the number of particles) and that the energy
\(E\) is real (which it always must be). Thus ends the solution.
Because the second derivative of the entropy is always
negative, the first derivative is monotonic, which means that the
temperature (which is positive) will always increase if you increase
the energy of the system and vice versa.
Extensive Internal Energy
S1 5455S
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 5455S
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 5455S
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.
