RMS stands for rootmeansquared, not relative molecular speed, so it's the square root of the mean of the squared velocities.JoeyC wrote:RMS (relative molecular speed0 =sqrt(3RT/M)
where R is the ideal gas constant, T is temperature, and M is mass.
Thermodynamics B/C

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Re: Thermodynamics B/C
I might be wrong, but doesn't RMS stand for root mean square?JoeyC wrote:RMS (relative molecular speed0 =sqrt(3RT/M)
where R is the ideal gas constant, T is temperature, and M is mass.
EDIT: wec01 beat me to it lol
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Re: Thermodynamics B/C
huh, whoops. Still the same equation though.
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Re: Thermodynamics B/C
Since nobody seems to be posting, here's a question:
Suppose there is a system containing 30 atoms that have 3 accessible microstates.
a) How many atoms must be in each microstate in order for entropy to be maximized?
b) What is the entropy of the system in part a)?
Suppose there is a system containing 30 atoms that have 3 accessible microstates.
a) How many atoms must be in each microstate in order for entropy to be maximized?
b) What is the entropy of the system in part a)?
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Re: Thermodynamics B/C
wec01 wrote:Since nobody seems to be posting, here's a question:
Suppose there is a system containing 30 atoms that have 3 accessible microstates.
a) How many atoms must be in each microstate 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]

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Re: Thermodynamics B/C
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 microstates.
a) How many atoms must be in each microstate 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]
You are using the correct formula however omega doesn't represent the number of microstates 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
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Re: Thermodynamics B/C
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?
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?

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Re: Thermodynamics B/C
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^{\gamma1}}=1\frac{1}{10^{\frac23}}=0.784[/math] (or 0.8 with sig figs) [math]\eta_{Diesel}=1\frac{\alpha^\gamma1}{r^{\gamma1}\gamma(\alpha1)}=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^\gamma1}{\gamma(\alpha1)}=1\implies\alpha^\gamma1=\gamma(\alpha1)[/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.
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Re: Thermodynamics B/C
Correct!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?If [math]r[/math] is compression ratio and [math]\alpha[/math] is cutoff ratio: [math]\eta_{Otto}=1\frac{1}{r^{\gamma1}}=1\frac{1}{10^{\frac23}}=0.784[/math] (or 0.8 with sig figs) [math]\eta_{Diesel}=1\frac{\alpha^\gamma1}{r^{\gamma1}\gamma(\alpha1)}=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^\gamma1}{\gamma(\alpha1)}=1\implies\alpha^\gamma1=\gamma(\alpha1)[/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.

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Re: Thermodynamics B/C
What type of heat engine can a hurricane be likened to? What is its source of heat and where does it exhaust the heat to? Assuming typical temperatures, pressures, etc. for the source of heat and exhaust (you may want to look up these typical values), what is its theoretical maximum efficiency?
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