Results: entropy

grading Quiz

60 min.

Free expansion

adiabatic expansion entropy temperature ideal gas

Students will determine the change in entropy (positive, negative, or none) for both the system and surroundings in three different cases. This is followed by an active whole-class discussion about where the entropy comes from during an irreversible process.

assignment Homework

Nucleus in a Magnetic Field

Nuclei of a particular isotope species contained in a crystal have spin \(I=1\), and thus, \(m = \{+1,0,-1\}\). The interaction between the nuclear quadrupole moment and the gradient of the crystalline electric field produces a situation where the nucleus has the same energy, \(E=\varepsilon\), in the state \(m=+1\) and the state \(m=-1\), compared with an energy \(E=0\) in the state \(m=0\), i.e. each nucleus can be in one of 3 states, two of which have energy \(E=\varepsilon\) and one has energy \(E=0\).

  1. Find the Helmholtz free energy \(F = U-TS\) for a crystal containing \(N\) nuclei which do not interact with each other.

  2. Find an expression for the entropy as a function of temperature for this system. (Hint: use results of part a.)

  3. Indicate what your results predict for the entropy at the extremes of very high temperature and very low temperature.

assignment Homework

Free Expansion

The internal energy is of any ideal gas can be written as \begin{align} U &= U(T,N) \end{align} meaning that the internal energy depends only on the number of particles and the temperature, but not the volume.*

The ideal gas law \begin{align} pV &= Nk_BT \end{align} defines the relationship between \(p\), \(V\) and \(T\). You may take the number of molecules \(N\) to be constant. Consider the free adiabatic expansion of an ideal gas to twice its volume. “Free expansion” means that no work is done, but also that the process is also neither quasistatic nor reversible.
  1. What is the change in entropy of the gas? How do you know this?

  2. What is the change in temperature of the gas?

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