Thermodynamics B/C

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Re: Thermodynamics B/C

Post by wec01 » February 22nd, 2019, 4:12 pm

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

Post by smayya337 » February 22nd, 2019, 4:12 pm

JoeyC wrote:RMS (relative molecular speed0 =sqrt(3RT/M)
where R is the ideal gas constant, T is temperature, and M is mass.
I might be wrong, but doesn't RMS stand for root mean square?

EDIT: wec01 beat me to it lol
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Re: Thermodynamics B/C

Post by JoeyC » February 22nd, 2019, 5:57 pm

huh, whoops. Still the same equation though.
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Re: Thermodynamics B/C

Post by wec01 » February 25th, 2019, 7:51 pm

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)?
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Re: Thermodynamics B/C

Post by smayya337 » February 26th, 2019, 3:35 pm

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]
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Re: Thermodynamics B/C

Post by wec01 » February 26th, 2019, 3:58 pm

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)?
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
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Re: Thermodynamics B/C

Post by smayya337 » February 27th, 2019, 1:21 pm

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?
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Re: Thermodynamics B/C

Post by mjcox2000 » February 27th, 2019, 3:37 pm

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.
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Re: Thermodynamics B/C

Post by smayya337 » February 27th, 2019, 4:27 pm

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^{\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.
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Re: Thermodynamics B/C

Post by mjcox2000 » February 27th, 2019, 8:12 pm

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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