Energy and Entropy: Fall-2024 Practice midterm (not graded) (SOLUTION): Due Day 14 Mon 10/14
Midterm practice ph423 piston
S1 5074S
Consider the pictured insulated piston, which is initially in thermal and mechanical equilibrium. The piston contains an unknown gas. You gradually add more mass to the top of the piston (perhaps by slowly pouring sand).
Does the pressure on the fluid increase, decrease, or remain the same, or can you not tell? Why?
Use the definition of pressure. Because the force increases, the pressure must also increase,
since the area of the piston remains the same.
Does the entropy of the fluid increase, decrease, or remain the same, or can you not tell? Why?
Use the relationship between entropy and energy transferred by heating. Because the
piston is insulated, there is no heating, and since the process is presumed to be
reasonably slow when I add a few masses, the entropy should not change.
Does the internal energy of the fluid increase, decrease, or remain the same, or can you not tell? Why?
Use the First Law. The masses do work on the system, and since there is no heating,
the internal energy must increase. We know that the masses will do work because the
gas will compress, which means the potential energy of the masses will decrease (i.e. they
lose energy). Or alternatively you could determine this algebraically by using \({\mathit{\unicode{273}}} W = -pdV\) and since \(dV<0\)
the work must be positive, meaning the system gains energy by working.
Midterm practice ph423 sensecheck
S1 5074S
Consider the following equations for internal energy and identify any problems that might indicate that they are erroneous. If the equation must be incorrect, please identify why it must be incorrect.
\(U = p S + \frac32 Nk_BT\)
Check dimensions and whether extensive is added to extensive. In this case extensivity works
out because the only extensive variables are \(U\), \(S\) and \(N\), which makes each of the three
terms also extensive (\(U\), \(pS\), and \( \frac32 Nk_BT\)).
When we check dimensions,
each term must be an energy, since \(U\) is an energy. Pressure is a force per area,
and entropy is an energy per temperature (as you can tell since \({\mathit{\unicode{273}}} Q=TdS\)).
These multiplied will not give us an energy, so the equation must be wrong.
\(U = \frac52 p V\)
Check dimensions and whether extensive is equal to extensive. Both sides are
extensive, and both sides are energies. You can remember that \(pV\) is an energy
because \({\mathit{\unicode{273}}} W = -pdV\).
\(U = \frac32Nk_BT\ln\left(1+N\right)\)
Check dimensions and whether extensive is added to extensive.
This one is a bit trickier. Both sides are energies, because \(k_BT\) is an energy
and \(N\) is dimensionless. So our dimensions work out fine.
The tricky bit here is to examine extensivity. \(U=Nk_BT\) works fine for extensivity, since
\(U\) and \(N\) are both extensive. What causes us trouble (and makes this wrong) is \(1+N\), since
the number 1 is not extensive and thus shouldn't be added to \(N\).
Consider the variable \(F\) defined by
\begin{align}
F = U - TS
\end{align}
where \(U\) is the internal energy, \(T\) is the temperature, and \(S\) is the entropy. Solve for the partial derivative \(\left(\frac{\partial F}{\partial T}\right)_V\) where \(V\) is the volume.
Hint: The partial derivative is defined by the ratio of small changes in \(F\) and \(T\) that would take us to a new state without changing volume. Since the changes are small, we can assume the change happens via a quasi-static process. Therefore, we know that the thermodynamic identity will be valid, \(dU = TdS - pdV\).
See the coffee and bagel problem for a similar problem.
We can zap this with a \(d\) to find that
\begin{align}
dF &= dU - TdS - SdT
\end{align}
Keep in mind that, when you zap with \(d\), anything can change, if it is not
absolutely and always a constant.
Now we need to eliminate the \(dU\), which we can do because we know the thermodynamic
identity.
\begin{align}
dF &= \left(\cancel{TdS}-pdV\right) - \cancel{TdS} - SdT \\
&= -SdT-pdV
\end{align}
By comparing this to the multivariable chain rule:
\begin{align}
dF=\left(\frac{\partial F}{\partial T}\right)_V\, dT+\left(\frac{\partial F}{\partial V}\right)_T\, dV
\end{align}
we can read off the derivative we want:
\begin{align}
\left(\frac{\partial F}{\partial T}\right)_V &= -S
\end{align}
