How do you solve for the final temperature in this insulated vessels problem?

[Erenjaeger]

Homework Statement

Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed as shown in the figure below. One vessel of volume V1 = 15.1 L,contains oxygen at a temperature of T1 = 320 K and a pressure of P1 = 1.70 atm. The other vessel of volume V2 = 22.5 L contains oxygen at a temperature of T2 = 445 K and a pressure of P2 = 1.90 atm. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout.

pic link : https://scontent-syd1-1.xx.fbcdn.ne...=4162bc2084f4f11ae61a2e1ad7058f8f&oe=572F8C7C

Can someone pls explain to me how you would go about solving this question, I have looked through my physics textbook but finding it hard to know where to start with this question.
Thanks.[/B]

 

[Erenjaeger]

any1 ^-^ pls

 

[TSny]

Hi, Erenjaeger!

Probably you are not getting any replies because you did not fill out parts 2 and 3 of the template. This is required for posting in the homework sections. You must show some attempt at a solution or at least some thoughts on what concepts are relevant.

If you add this additional material, I think you will get some response.

 

[Chestermiller]

How many moles of O2 are in V1 and in V2 initially?

 

[Erenjaeger]

How many moles of O2 are in V1 and in V2 initially?

I used PV=nRT, or n=PV/RT to find the number of moles in each vessel and to find the final pressure wouldn't I just use P(final)=n(of both vessels)xR(8.314J.K)xT/V(of both vessels)? If so, then I am stuck because I am unsure of how i would find the final temperature??
Also is this an Isovolumetric process since there is no net change in volume? or is there a net change when the two volumes combine ?

 

[Chestermiller]

I used PV=nRT, or n=PV/RT to find the number of moles in each vessel and to find the final pressure wouldn't I just use P(final)=n(of both vessels)xR(8.314J.K)xT/V(of both vessels)?

Yes. Very Excellent.

If so, then I am stuck because I am unsure of how i would find the final temperature??
Also is this an Isovolumetric process since there is no net change in volume? or is there a net change when the two volumes combine ?

If you consider the contents of the two vessels as your system, how much work does this system do on its surroundings (the vessels)? How much heat is transferred into or out of this insulated system? What is the change in internal energy of the system? What is the equation for the internal energy of an ideal gas?

 

[Erenjaeger]

Yes. Very Excellent.

If you consider the contents of the two vessels as your system, how much work does this system do on its surroundings (the vessels)? How much heat is transferred into or out of this insulated system? What is the change in internal energy of the system? What is the equation for the internal energy of an ideal gas?

Is any work done? because if the process is isovolumetric work is dependent upon change in volume and since there is no net change in the volume wouldn't that mean there is no work done? and the system is thermally insulated so wouldn't that imply that no heat is transferred out of the system? First law of thermodynamics: a conservation of energy statement. A change in the internal energy of a system of an ideal gas, ΔU, is due to heat transfer Q, in or out of the system and work, W, done on or by the system. The equation is ΔU=Q+w.
So there must be some work done or heat transfer or even both right?
Not too sure where the work or heat transfer would come from tho?? Maybe heat transfer would be from one ideal gas to another? since one is colder?

 

[Chestermiller]

Is any work done? because if the process is isovolumetric work is dependent upon change in volume and since there is no net change in the volume wouldn't that mean there is no work done?

Correct

and the system is thermally insulated so wouldn't that imply that no heat is transferred out of the system?

Correct.

First law of thermodynamics: a conservation of energy statement. A change in the internal energy of a system of an ideal gas, ΔU, is due to heat transfer Q, in or out of the system and work, W, done on or by the system. The equation is ΔU=Q+w.
So there must be some work done or heat transfer or even both right?

No. You were correct in concluding that $\Delta U = 0$

Not too sure where the work or heat transfer would come from tho?? Maybe heat transfer would be from one ideal gas to another? since one is colder?

Heat transfer from one gas to the other is not heat transferred into or out of our system. It is internal to the system.

If we take the internal energy of oxygen to be zero at an arbitrary reference temperature $T_r$, the internal energy of our system initially is $$U_{init}=n_1C_v(320 - T_r)+n_2C_v(445-T_r)$$where $C_v$ is the molar heat capacity of oxygen. If $T_f$ is the final temperature of our system, what is the corresponding equation for $U_{final}$ in terms of $T_f$, $n_1$, $n_2$, $C_v$, and $T_r$. From these two equations, what is the equation for $\Delta U=U_{final}-U_{initial}$?

 

[Erenjaeger]

If we take the internal energy of oxygen to be zero at an arbitrary reference temperature $T_r$, the internal energy of our system initially is $$U_{init}=n_1C_v(320 - T_r)+n_2C_v(445-T_r)$$where $C_v$ is the molar heat capacity of oxygen. If $T_f$ is the final temperature of our system, what is the corresponding equation for $U_{final}$ in terms of $T_f$, $n_1$, $n_2$, $C_v$, and $T_r$. From these two equations, what is the equation for $\Delta U=U_{final}-U_{initial}$?[/QUOTE]

I don't get what you mean here sorry ??

 

[Erenjaeger]

Correct

Correct.

No. You were correct in concluding that $\Delta U = 0$

Heat transfer from one gas to the other is not heat transferred into or out of our system. It is internal to the system.

If we take the internal energy of oxygen to be zero at an arbitrary reference temperature $T_r$, the internal energy of our system initially is $$U_{init}=n_1C_v(320 - T_r)+n_2C_v(445-T_r)$$where $C_v$ is the molar heat capacity of oxygen. If $T_f$ is the final temperature of our system, what is the corresponding equation for $U_{final}$ in terms of $T_f$, $n_1$, $n_2$, $C_v$, and $T_r$. From these two equations, what is the equation for $\Delta U=U_{final}-U_{initial}$?

I don't understand what you mean here sorry??

 

[Chestermiller]

I don't understand what you mean here sorry??

I'm assuming that you are aware that the internal energy of an ideal gas is a function only of temperature, and that dU=nC_vdT.

If we integrate this equation between the arbitrary reference temperature T_r (taken as the datum for zero internal energy) and the initial temperatures of the gases in the two vessels, we can get the internal energy of each of the gases, and we can then add them together to get the total initial internal energy of gases within the two vessels. This led to the equation for the initial internal energy $$U_{init}=n_1C_v(320 - T_r)+n_2C_v(445-T_r)$$
By the same rationale, we can get the final internal energy of the system as:
$$U_{final}=(n_1+n_2)C_v(T_f-T_r)$$where $T_f$ is the final temperature of the system. So, the change in internal energy is obtained by taking the difference between the initial and final internal energies:
$$\Delta U=U_{final}-U_{initial}=n_1C_v(T_f-320)+n_2C_v(T_f-445)$$
Note that the reference temperature T_r no longer appears in this equation.

We know that the change in internal energy of the system has to be equal to zero. From the above equation, what does this give for the final temperature?

 

[Erenjaeger]

I'm assuming that you are aware that the internal energy of an ideal gas is a function only of temperature, and that dU=nC_vdT.

If we integrate this equation between the arbitrary reference temperature T_r (taken as the datum for zero internal energy) and the initial temperatures of the gases in the two vessels, we can get the internal energy of each of the gases, and we can then add them together to get the total initial internal energy of gases within the two vessels. This led to the equation for the initial internal energy $$U_{init}=n_1C_v(320 - T_r)+n_2C_v(445-T_r)$$
By the same rationale, we can get the final internal energy of the system as:
$$U_{final}=(n_1+n_2)C_v(T_f-T_r)$$where $T_f$ is the final temperature of the system. So, the change in internal energy is obtained by taking the difference between the initial and final internal energies:
$$\Delta U=U_{final}-U_{initial}=n_1C_v(T_f-320)+n_2C_v(T_f-445)$$
Note that the reference temperature T_r no longer appears in this equation.

We know that the change in internal energy of the system has to be equal to zero. From the above equation, what does this give for the final temperature?

Okay, so
using that information i have: ΔU=979.35x37.6x(T_f-320)+1170.80x37.6x(T_f-445)

using C_v as 37.6 because that's the sum of the two volumes, am i correct in using that value for C_v ??
so that equation has to equal 0 right?
If so I am not really sure how i can solve for T_f ? assuming that is what I am meant to be doing right?
Aslo i just wanted to say thanks, its crazy how helpful and patient people are on this site :D

update: MY ATTEMPT,
so i expanded the equation
then got all the T_f on one side and just solved that equation for T_f
and i got T_f = 29.17 °C ??

Update: I solved the equation for T_f again and got 388.063K (which is right)
Thank you

 

[Chestermiller]

Okay, so
using that information i have: ΔU=979.35x37.6x(T_f-320)+1170.80x37.6x(T_f-445)

using C_v as 37.6 because that's the sum of the two volumes, am i correct in using that value for C_v ??

You are aware that C_v cancels out, correct?

so that equation has to equal 0 right?

Right.

If so I am not really sure how i can solve for T_f ? assuming that is what I am meant to be doing right?
Aslo i just wanted to say thanks, its crazy how helpful and patient people are on this site :D

update: MY ATTEMPT,
so i expanded the equation
then got all the T_f on one side and just solved that equation for T_f
and i got T_f = 29.17 °C ??

Update: I solved the equation for T_f again and got 388.063K (which is right)
Thank you

What did you get for n1 and n2? (The 388 might be correct, but the 29 is definitely incorrect)

 

[Erenjaeger]

You are aware that C_v cancels out, correct?

Right.

What did you get for n1 and n2? (The 388 might be correct, but the 29 is definitely incorrect)

I got by expanding the equation then putting all the T_f terms on one side and then solved for T_f² and just sqrt the answer for T_f which maybe i did wrong to get 29, and then i tried again. First i refined the equation then multiplied both sides by 100, then expanded the equation 3137336480 to both sides, and then divided both sides by 8084564 and got x=16687960/43003
which is x=388.063K

 

[Erenjaeger]

I got by expanding the equation then putting all the T_f terms on one side and then solved for T_f² and just sqrt the answer for T_f which maybe i did wrong to get 29, and then i tried again. First i refined the equation then multiplied both sides by 100, then expanded the equation 3137336480 to both sides, and then divided both sides by 8084564 and got x=16687960/43003
which is x=388.063K

yeah i thought it did cancel out, which i did cancel it out during that working above

 

[Chestermiller]

yeah i thought it did cancel out, which i did cancel it out during that working above

How and why did you end up with a $T_f^2$, since you started out with a simple linear algebraic equation in $T_f$?

For $n_1$ and $n_2$, I got $$n_1=\frac{PV}{RT}=\frac{(15.1)(1.7)}{(0.082)(320)}=0.978$$
$$n_2=\frac{PV}{RT}=\frac{(22.5)(1.9)}{0.082)(445)}=1.172$$

Both of these are in units of gram moles.

The linear algebraic equation that $T_f$ satisfies is:

$$\Delta U=U_{final}-U_{initial}=n_1C_v(T_f-320)+n_2C_v(T_f-445)=0$$

Solving algebraically for $T_f$ gives:
$$T_f=\frac{320n_1+445n_2}{(n_1+n_2)}=388.1$$

 

[Erenjaeger]

How and why did you end up with a $T_f^2$, since you started out with a simple linear algebraic equation in $T_f$?

For $n_1$ and $n_2$, I got $$n_1=\frac{PV}{RT}=\frac{(15.1)(1.7)}{(0.082)(320)}=0.978$$
$$n_2=\frac{PV}{RT}=\frac{(22.5)(1.9)}{0.082)(445)}=1.172$$

Both of these are in units of gram moles.

The linear algebraic equation that $T_f$ satisfies is:

$$\Delta U=U_{final}-U_{initial}=n_1C_v(T_f-320)+n_2C_v(T_f-445)=0$$

Solving algebraically for $T_f$ gives:
$$T_f=\frac{320n_1+445n_2}{(n_1+n_2)}=388.1$$

I ended up with that T_f² because i tried to solve the equation without cancelling out the C_v and moving both of the T_f to one side which i thought would just be T_f ² i think that's why i got the wrong answer initially. for n₁ =PV/RT n₁ = (172252x15.1)/(8.314x320) n₁ = 979.35 moles and for n₂ = PV/RT n₂ = (192518x22.5)/(8.314x445) n₂ = 1170.80 moles
and used that info for to solve the equation cancelling out C_v then I refined the equation then multiplied both sides by 100, then expanded the equation then added 3137336480 to both sides, and then divided both sides by 8084564 and got x=16687960/43003
which is x=388.063K

 

[Chestermiller]

I ended up with that T_f² because i tried to solve the equation without cancelling out the C_v and moving both of the T_f to one side which i thought would just be T_f ² i think that's why i got the wrong answer initially. for n₁ =PV/RT n₁ = (172252x15.1)/(8.314x320) n₁ = 979.35 moles and for n₂ = PV/RT n₂ = (192518x22.5)/(8.314x445) n₂ = 1170.80 moles
and used that info for to solve the equation cancelling out C_v then I refined the equation then multiplied both sides by 100, then expanded the equation then added 3137336480 to both sides, and then divided both sides by 8084564 and got x=16687960/43003
which is x=388.063K

Each of your number of moles is off by a factor of 1000.