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JoeyC wrote:RMS (relative molecular speed0 =sqrt(3RT/M)
where R is the ideal gas constant, T is temperature, and M is mass.

JoeyC wrote:RMS (relative molecular speed0 =sqrt(3RT/M)
where R is the ideal gas constant, T is temperature, and M is mass.

wec01 wrote:Since nobody seems to be posting, here's a question:

Suppose there is a system containing 30 atoms that have 3 accessible micro-states.

a) How many atoms must be in each micro-state in order for entropy to be maximized?

b) What is the entropy of the system in part a)?

a) With maximum entropy, all microstates are equally probable, so the number of atoms in each should be 30/3 = [b]10.[/b]
b) The formula for entropy is S = k ln(W), so we can plug in Boltzmann's constant of 1.38065 x 10^-23 J/K and the number of microstates into k and W, respectively, to get [b]S = 1.38065 * 10^(-23) * ln(3) = 1.5227316 * 10^-23 J/K.[/b]

smayya337 wrote:

wec01 wrote:Since nobody seems to be posting, here's a question:

Suppose there is a system containing 30 atoms that have 3 accessible micro-states.

a) How many atoms must be in each micro-state in order for entropy to be maximized?

b) What is the entropy of the system in part a)?

I had to look this one up and I'm probably still wrong (Click to Show Hidden Text)

a) With maximum entropy, all microstates are equally probable, so the number of atoms in each should be 30/3 = [b]10.[/b]
b) The formula for entropy is S = k ln(W), so we can plug in Boltzmann's constant of 1.38065 x 10^-23 J/K and the number of microstates into k and W, respectively, to get [b]S = 1.38065 * 10^(-23) * ln(3) = 1.5227316 * 10^-23 J/K.[/b]

You are using the correct formula however omega doesn't represent the number of micro-states it represents the number of ways the system's state can be achieved. In this case it would be the number of combinations that have 10 atoms in each state so: [math]30!/(10!)^3[/math] which would give about 4.05 * 10^-22 J/K

smayya337 wrote:An engine has a compression ratio of 10. The gas inside is an ideal monatomic gas.

a) Calculate the Otto efficiency of this engine.
b) Calculate the Diesel efficiency of this engine with a cutoff ratio of 5.
c) Is there a value for the cutoff ratio that causes the Otto and Diesel efficiencies to be equal? If so, what is it?

If [math]r[/math] is compression ratio and [math]\alpha[/math] is cutoff ratio:
[math]\eta_{Otto}=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{10^{\frac23}}=0.784[/math] (or 0.8 with sig figs)
[math]\eta_{Diesel}=1-\frac{\alpha^\gamma-1}{r^{\gamma-1}\gamma(\alpha-1)}=1-\frac{5^{\frac53}-1}{10^{\frac23}\cdot\frac53\cdot4}=0.559[/math] (or 0.6 with sig figs)
These efficiencies are equal when [math]\frac{\alpha^\gamma-1}{\gamma(\alpha-1)}=1\implies\alpha^\gamma-1=\gamma(\alpha-1)[/math], which is satisfied if [math]\alpha=1[/math]. However, this has no physical significance as [math]\alpha=1[/math] would correspond to an engine doing no work per cycle.

mjcox2000 wrote:

smayya337 wrote:An engine has a compression ratio of 10. The gas inside is an ideal monatomic gas.

a) Calculate the Otto efficiency of this engine.
b) Calculate the Diesel efficiency of this engine with a cutoff ratio of 5.
c) Is there a value for the cutoff ratio that causes the Otto and Diesel efficiencies to be equal? If so, what is it?

Answer (Click to Show Hidden Text)

If [math]r[/math] is compression ratio and [math]\alpha[/math] is cutoff ratio:
[math]\eta_{Otto}=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{10^{\frac23}}=0.784[/math] (or 0.8 with sig figs)
[math]\eta_{Diesel}=1-\frac{\alpha^\gamma-1}{r^{\gamma-1}\gamma(\alpha-1)}=1-\frac{5^{\frac53}-1}{10^{\frac23}\cdot\frac53\cdot4}=0.559[/math] (or 0.6 with sig figs)
These efficiencies are equal when [math]\frac{\alpha^\gamma-1}{\gamma(\alpha-1)}=1\implies\alpha^\gamma-1=\gamma(\alpha-1)[/math], which is satisfied if [math]\alpha=1[/math]. However, this has no physical significance as [math]\alpha=1[/math] would correspond to an engine doing no work per cycle.