Thermodynamics study questions

[dcrisci]

THIS WAS MOVED FROM ANOTHER FORUM, SO THERE IS NO TEMPLATE. HOWEVER, THE OP DID SHOW SOME EFFORT

Hello, I have been having troubles beginning these two problems given for exam prep. Was wondering if anyone could give guidance on where to begin.

Problem 1:
A cylinder with adiabatically isolated walls is closed at both ends and is divided into two volumes by a frictionless piston that is also thermally insulating. Initially, the volume, temperature and pressure of the ideal gas on each side of the cylinder are equal at V0, T0 and p0, respectively. A heating coil, inserted in the right hand volume, is used to heat slowly the gas on that side until the pressure reaches (64/27) p0 = 2.37 p0. If the heat capacity CV of the gas is independent of temperature, and Cp/CV = 1.5,
(a) find first the entropy change of the gas on the left (no calculation needed for this step!),
(b) then calculate the final left–hand volume in terms of V0,
(c) and finally, calculate the final temperature in terms of T0 on the left hand side.

For this problem, I am unsure of (b). For (a) I determined that since the system is adiabatically isolated, and the piston is also thermally insulating, no heat can enter or exit the volume on the left. Because of this the change in entropy is equal to zero since Q = 0.
(b) is slightly confusing for me since I don't where to begin. If we heat up the RHS with the heating coil and then shut it off when pressure reaches 2.37p0, we are unsure if the volume has remained the same while the temperature and pressure increase.
I figure part (c) requires part (b) to solve so I haven't gotten this far.

Problem 2:
A column of water contains fine metal particles of radius 20 nm, which are in thermal equilibrium at 25°C. If there are 1000 such particles per unit volume at a given height h0 in the water column, how many particles would be found in the same volume 1.0 mm higher than h0? The density of the used metal is ρ = 20.0 g/cm3.

For this problem I have not made any progress on it as I have no clue what equations pertain to this problem and was hoping someone could lead me in the right direction.

Thanks everyone!

 

[Chestermiller]

THIS WAS MOVED FROM ANOTHER FORUM, SO THERE IS NO TEMPLATE. HOWEVER, THE OP DID SHOW SOME EFFORT

Hello, I have been having troubles beginning these two problems given for exam prep. Was wondering if anyone could give guidance on where to begin.

PLEASE MOVE PROBLEM 2 TO A NEW THREAD. WE SHOULD NOT POST TWO PROBLEMS IN THE SAME THREAD. THIS CREATES GREAT CONFUSION. AFTER YOU START THE NEW THREAD FOR PROBLEM 2, I WILL EDIT IT OUT OF THIS THREAD.

Problem 1:
A cylinder with adiabatically isolated walls is closed at both ends and is divided into two volumes by a frictionless piston that is also thermally insulating. Initially, the volume, temperature and pressure of the ideal gas on each side of the cylinder are equal at V0, T0 and p0, respectively. A heating coil, inserted in the right hand volume, is used to heat slowly the gas on that side until the pressure reaches (64/27) p0 = 2.37 p0. If the heat capacity CV of the gas is independent of temperature, and Cp/CV = 1.5,
(a) find first the entropy change of the gas on the left (no calculation needed for this step!),
(b) then calculate the final left–hand volume in terms of V0,
(c) and finally, calculate the final temperature in terms of T0 on the left hand side.

For this problem, I am unsure of (b). For (a) I determined that since the system is adiabatically isolated, and the piston is also thermally insulating, no heat can enter or exit the volume on the left. Because of this the change in entropy is equal to zero since Q = 0.

This is correct.

(b) is slightly confusing for me since I don't where to begin. If we heat up the RHS with the heating coil and then shut it off when pressure reaches 2.37p0, we are unsure if the volume has remained the same while the temperature and pressure increase.

How do the pressures on the two sides compare when the system reaches the final state? What is the relationship between pressure and volume in the left chamber which, as you have already pointed out, is experiencing an adiabatic reversible compression?

 

[dcrisci]

How do the pressures on the two sides compare when the system reaches the final state? What is the relationship between pressure and volume in the left chamber which, as you have already pointed out, is experiencing an adiabatic reversible compression?

So what I have gotten is that P₀V₀^γ = P_fV_f^γ

And from that I have found that the volume on the RHS (with the heating coil) has a final volume of $\frac{V_0}{2.37^\frac{2}{3}}$
Does this seem correct?
Then with this value I used the fact that the cylinder has a volume of $2 V_0$ and then the volume of the LHS will be the difference between these two.
$V_{LHS} = 2 V_0 - \frac{V_0}{2.37^\frac{2}{3}} = 1.437 V_0$

Also for the temperature on the LHS I used the equation:
V₀T₀^γ = V_fT_f^γ
Rearranged it for the final temperature to get
$T_f = T_0 [\frac{V_f}{V_0}]^\frac{1}{γ}$
Which have an answer of
$T_f = 0.785 T_0$
Note the value of γ that I used was 3/2

 

[Chestermiller]

So what I have gotten is that P₀V₀^γ = P_fV_f^γ

And from that I have found that the volume on the RHS (with the heating coil) has a final volume of $\frac{V_0}{2.37^\frac{2}{3}}$
Does this seem correct?

Yes.
There was a reason they also expressed it as 64/27. What is 64/27 raised to the 2/3 power (as a proper fraction)? I get 16/9.

Then with this value I used the fact that the cylinder has a volume of $2 V_0$ and then the volume of the LHS will be the difference between these two.
$V_{LHS} = 2 V_0 - \frac{V_0}{2.37^\frac{2}{3}} = 1.437 V_0$

You already determined the volume of the left hand chamber as 0.5625 V₀. The calculation you have shown here would be for the volume of the right hand chamber.

Also for the temperature on the LHS I used the equation:
V₀T₀^γ = V_fT_f^γ
Rearranged it for the final temperature to get
$T_f = T_0 [\frac{V_f}{V_0}]^\frac{1}{γ}$
Which have an answer of
$T_f = 0.785 T_0$
Note the value of γ that I used was 3/2

Once you knew the pressure and volume of the left chamber, you could have used the ideal gas law to determine the temperature. I get (4/3)T₀.

Actually, the only mistake you made was using the volume of the right hand chamber in the calculation.

 

[dcrisci]

Yes.
There was a reason they also expressed it as 64/27. What is 64/27 raised to the 2/3 power (as a proper fraction)? I get 16/9.
You already determined the volume of the left hand chamber as 0.5625 V₀. The calculation you have shown here would be for the volume of the right hand chamber.Once you knew the pressure and volume of the left chamber, you could have used the ideal gas law to determine the temperature. I get (4/3)T₀.

Actually, the only mistake you made was using the volume of the right hand chamber in the calculation.

How would it be the volume of the left hand chamber being 0.5625 V₀, It says in the question that the heating coil is in the right hand side and is in it until the pressure is 2.37 P₀.
"A heating coil, inserted in the right hand volume, is used to heat slowly the gas on that side until the pressure reaches (64/27) p0 = 2.37 p0"
So if the right hand side is at this pressure, and we use the equation I gave above, wouldn't we be finding the final volume on the right hand side?

 

[Chestermiller]

How would it be the volume of the left hand chamber being 0.5625 V₀, It says in the question that the heating coil is in the right hand side and is in it until the pressure is 2.37 P₀.

At equilibrium, the pressures in the two chambers have to be equal. Otherwise, the piston won't be in force equilibrium.

"A heating coil, inserted in the right hand volume, is used to heat slowly the gas on that side until the pressure reaches (64/27) p0 = 2.37 p0"
So if the right hand side is at this pressure, and we use the equation I gave above, wouldn't we be finding the final volume on the right hand side?

No. The right chamber is not adiabatic.

 

[dcrisci]

At equilibrium, the pressures in the two chambers have to be equal. Otherwise, the piston won't be in force equilibrium.No. The right chamber is not adiabatic.

So the question is really saying that the heating coil is in the RHS until the pressure in the LHS is 2.37 P₀, and that's the side we work with because Q = 0 on that side?
I always did think that the volume on the RHS (chamber with heating coil) was supposed to increase, but the question is worded weird and difficult to understand I guess.

 

[Chestermiller]

So the question is really saying that the heating coil is in the RHS until the pressure in the LHS is 2.37 P₀, and that's the side we work with because Q = 0 on that side?

Yes. But note that the 2.37 P₀ is the final pressure in both chambers.

I always did think that the volume on the RHS (chamber with heating coil) was supposed to increase, but the question is worded weird and difficult to understand I guess.

It didn't seem confusing to me. The focus of this problem was the left chamber.

 

[dcrisci]

Yes. But note that the 2.37 P₀ is the final pressure in both chambers.

OHHH that makes much more sense to me now! Thank you!
I will post a solution of this soon enough