Thermodynamics final temperature of gas

In summary: O2 gas is divided up between translational and rotational KE. This means that the heat capacity of O2 is 5R/2, while that of He is 3R/2. So you must use the correct heat capacity according to the gas that you are using. The heat capacity is different for different gases. The teacher's method is correct because of the law of conservation of energy. Energy is conserved in the system, so the total energy before the valve is opened must be equal to the total energy after the valve is opened. Since the gases are closed systems, there is no external work being done, so the only form of energy that can change is heat. Therefore, the heat
  • #1
Legendon
22
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Homework Statement


Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed. One vessel of volume 16.8 L contains oxygen at a temperature of 300 K and a pressure of 1.75 atm. The other vessel of volume 22.4 L contains oxygen at a temperature of 450 K and a pressure of 2.25 atm. When the valve is opened, the gases in the two vessels mix, and the temperature and pressure become uniform throughout. (a) What is the final temperature? (b) What is the final pressure?
(Ans. (a) 380 K , (b) 2.06 x 105 Pa or 2.04 atm)


Homework Equations





The Attempt at a Solution


The teacher used the method of
heat in=heat out
(m1)c(T-300)=(m2)c(450-T)
and obtain T=380K
I did it by
Total Energy of system is conserved.
For an ideal gas, U=KE
3/2(n1)R300+3/2(n2)R450=3/2(n1+n2)RT
T=380K
Is the method i used correct?
And how come the teachers method is correct? The pressure is different in the vessels. So shouldn't that be taken into account?
 
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  • #2
well both the final temperature and final pressure of the two systems are the same at the end, which is the time you are calculating for.

you used conservation of energy, and the teacher used conservation of heat. There are two energy components to U (the total internal energy). Those two components are heat, and work. Work is only done if there is change in volume, but in this case the vessels do not grow or shrink. So the change in heat will be equal to the change in total internal energy will be equal to 0 for this case(thermally insulated). This change being 0 allows you to easily relate the final and initial energy values.

EDIT: To be more specific as to why your equations are the same, let me try to put it this way.
He used mass, you used moles. Those have a linear relationship with each other since it is oxygen that's in both tanks. You used R, he used c, both constants that will give the units of energy. Then you used temperature, and he used temperature. So as you can see, both formulas are quite similar.
 
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  • #3
Legendon said:

Homework Statement


Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed. One vessel of volume 16.8 L contains oxygen at a temperature of 300 K and a pressure of 1.75 atm. The other vessel of volume 22.4 L contains oxygen at a temperature of 450 K and a pressure of 2.25 atm. When the valve is opened, the gases in the two vessels mix, and the temperature and pressure become uniform throughout. (a) What is the final temperature? (b) What is the final pressure?
(Ans. (a) 380 K , (b) 2.06 x 105 Pa or 2.04 atm)

The Attempt at a Solution


The teacher used the method of
heat in=heat out
(m1)c(T-300)=(m2)c(450-T)
and obtain T=380K
I did it by
Total Energy of system is conserved.
For an ideal gas, U=KE
3/2(n1)R300+3/2(n2)R450=3/2(n1+n2)RT
T=380K
Is the method i used correct?
And how come the teachers method is correct? The pressure is different in the vessels. So shouldn't that be taken into account?
Your method is correct but you are using the incorrect heat capacity. [itex]O_{2}[/itex] is a diatomic gas and its heat capacity at constant volume is 5R/2 in this temperature range. But, since heat capacities cancel out it doesn't matter.

Your method is equivalent to your teacher's method. Your teacher is using the heat capacity per unit mass and you are using the molar heat capacity. Since the two gases are the same they all cancel out:

m1c(T-300) = m2c(450-T)
m1cT - m1c(300) = m2c(450) - m2cT
(m1+m2)T = m2(450)+m1(300)

AM
 
  • #4
I am not using molar heat capacity? on my notes it says Kinetic energy=3/2nRT??. But i am confused with the teacher's method. Is heat dependent on pressure? the teachers method is i see it as putting solution1 of temp1 and solution2 of temp2 together and finding final temp. But with gas of differernt initial pressures, this method still works without considering the pressures ?
 
  • #5
You're asking a good question. Yes the pressure and the heat are related, which I believe covers the missing link in your understanding. you have constant pressure, constant temperature, constant volume, and constant moles inside the whole system at its final time. You know what all of these values are, therefore you will be able to find the temperature. He didn't need to take into account pressure, just as he didn't take into account Volume because PV = nRT. He has nRT, just in the form of m.c.dT

Your teacher is saying that all of the heat lost by the hotter gas must be gained by the cooler gas. What he is doing is simply calculating the total heat inside the system, and given all of those constant parameters i mentioned in the first paragraph, he can find the final temperature.

His equation is very similar to your equation. It is like this:

(m1)c(T1) + (m2)c(T2) = (m1+m2)c(Tfinal)
 
  • #6
"He didn't need to take into account pressure, just as he didn't take into account Volume because PV = nRT. He has nRT, just in the form of m.c.dT." Thank you.
 
  • #7
Legendon said:
I am not using molar heat capacity? on my notes it says Kinetic energy=3/2nRT??. But i am confused with the teacher's method. Is heat dependent on pressure? the teachers method is i see it as putting solution1 of temp1 and solution2 of temp2 together and finding final temp. But with gas of differernt initial pressures, this method still works without considering the pressures ?
I thought you were using 3R/2 as the heat capacity. I see that you were using a kind of principle of conservation of translational kinetic energy. Unfortunately, there is no such principle. So your method will only work if the two gases have the same heat capacity, Cv.

For example, it will not work if one gas is He and the other is O2. The internal energy of He is stored as translational KE. However, O2 stores kinetic energy as rotational as well as translational KE. But only translational KE determines temperature. So for a given change in internal energy there will be a greater change in the temperature of He than of O2.

AM
 
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  • #8
Ok, today the lecturer started on kinetic theory of gasses, i realized jR/2 is the heat capacity. and for oxygen, j=5. Thanks.
 

FAQ: Thermodynamics final temperature of gas

1. What is the final temperature of a gas in a thermodynamics system?

The final temperature of a gas in a thermodynamics system is determined by the initial temperature, pressure, and volume of the gas, as well as any changes in these variables due to external factors such as heat transfer or work done on the system. It is calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

2. How can the final temperature of a gas in a thermodynamics system be calculated?

The final temperature of a gas in a thermodynamics system can be calculated using the ideal gas law, which states that the product of pressure and volume is directly proportional to the product of temperature and the number of moles of gas. This relationship can be rearranged to solve for the final temperature, given the initial conditions and any changes in pressure or volume.

3. What factors can affect the final temperature of a gas in a thermodynamics system?

The final temperature of a gas in a thermodynamics system can be affected by changes in pressure, volume, and heat transfer. Additionally, the type of gas and its specific heat capacity can also have an impact on the final temperature. Any external work done on the system, such as compression or expansion, can also affect the final temperature.

4. Can the final temperature of a gas in a thermodynamics system ever be lower than the initial temperature?

Yes, the final temperature of a gas in a thermodynamics system can be lower than the initial temperature. This can occur if the gas expands and does work on its surroundings, causing a decrease in internal energy and therefore a decrease in temperature. This is known as an adiabatic process, where there is no heat transfer between the system and its surroundings.

5. What is an isothermal process and how does it affect the final temperature of a gas in a thermodynamics system?

An isothermal process is one in which the temperature of the system remains constant throughout. In this case, the final temperature of a gas in a thermodynamics system will be equal to the initial temperature, as there is no change in internal energy. This can occur when the system is in thermal equilibrium with its surroundings, and there is no net heat transfer between the two.

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