Temperature of gas when internal energy is constant

[sgstudent]

Homework Statement

A fixed mass of gas is trapped in a metal cylinder by a movable piston. The piston is moved outwards slowly. The volume of gas increases but the internal energy is unchanged. What happens to the pressure and to the temperature of the gas?

Homework Equations

none

The Attempt at a Solution

I think the pressure and temperature will decrease because when I move the piston back, the distance apart each gas particle increases so the potential energy will increase (similar to boiling or melting when particle distance increases). So for internal energy to remain unchanged, the kinetic energy have to decrease explaining the decrease in temperature. So with the greater volume and lower temperature pressure has to be decreased.

Is this correct? This was an O level MCQ question to give everyone a better understanding of the level I am in.

Thanks for the help

 

[Saitama]

I am not the right person to help in Thermodynamics but internal energy is directly related to temperature. Doesn't that mean the temperature remains constant? I am not sure if this is the same for both ideal and real gas.

 

[sgstudent]

I am not the right person to help in Thermodynamics but internal energy is directly related to temperature. Doesn't that mean the temperature remains constant? I am not sure if this is the same for both ideal and real gas.

I have not learned about it but I read up in it. It assumes that potential energy is negligible right? But since potential energy indeed does increase so shouldn't kinetic energy decrease?

I mean its unfair to not accept my answer as it is actually more correct than suggested one.

 

[sankalpmittal]

I have not learned about it but I read up in it. It assumes that potential energy is negligible right? But since potential energy indeed does increase so shouldn't kinetic energy decrease?

I mean its unfair to not accept my answer as it is actually more correct than suggested one.

ΔU=0 , in isothermal process... (Another evidence that since process is taking place infinitesimally slowly , its reversible also.) And no , its all about internal energy , it will not be correct to consider potential energy in ideal gas. Think this the other way... ΔU = ΔK , for sure... As ΔU=0 , so ΔK= 0 , then what happens to temperature ? In isothermal process , gas will follow boyle's law which states that ...?

Once you answer , this you're done...

 

[Chestermiller]

It is important to mention that the internal energy is solely a function for temperature only in the ideal gas region. For the non-ideal gas region, the internal energy is also a function of pressure.

 

[sgstudent]

It is important to mention that the internal energy is solely a function for temperature only in the ideal gas region. For the non-ideal gas region, the internal energy is also a function of pressure.

But when considering real gases then won't the potential energy play a role. Also, under my O levels syllabus we weren't taught about ideal or real gas laws but we know that when the distance between particles increase (melting, boiling) the potential energy increases. So without any prior knowledge to those laws would my answer be more 'correct'. Since under the whole spectrum of physics both answers should theoratically be correct.

Thanks for the help!

 

[ehild]

In case of a solid or liquid, the particles are close and they interact. They are kept at their equilibrium distance by some forces mainly of electric nature, and the energy is minimum at that equilibrium position. The internal energy includes both kinetic and potential energy of the random motion of the particles about the equilibrium position, and it is determined by the temperature. But the distance between the particles of the solid or liquid would not change if you put them into a bigger container.

The particles of a normal gas are far enough from each other so that their interaction is negligible, except those very rare occasions when the atoms/molecules collide. The potential energy between the particles of gas can be ignored. That is why the particles of a gas occupy the whole volume of the container: There is no force keeping them back, except the walls of the vessel.

So you can accept that the internal energy of the gas at normal temperatures is equal to the kinetic energy of the random motion of its molecules, and that kinetic energy is determined by the temperature alone. If the internal energy is unchanged the temperature is also constant.

On the other hand, the same amount of gas is kept in the vessel, only the volume increases. The pressure is the average force/unit area the particles of the gas exert on the walls, and that depends on the speed of the particles and the number of collisions in unit time. The number of collisions decreases with greater volume. You certainly heard about Boyle's Law. http://en.wikipedia.org/wiki/Boyle's_law

ehild

 

[Chestermiller]

So you can accept that the internal energy of the gas at normal temperatures is equal to the kinetic energy of the random motion of its molecules, and that kinetic energy is determined by the temperature alone. If the internal energy is unchanged the temperature is also constant.

ehild

I stand by what I said. The internal energy of a real gas is, in general, a function of both temperature and pressure, and deviates from a pure dependence on temperature under conditions where the gas deviates from PV = nRT. The dependence of internal energy on pressure can be precisely calculated from the gas' equation of state.

 

[ehild]

Homework Statement

A fixed mass of gas is trapped in a metal cylinder by a movable piston. The piston is moved outwards slowly. The volume of gas increases but the internal energy is unchanged. What happens to the pressure and to the temperature of the gas?

Homework Equations

none

The Attempt at a Solution

I think the pressure and temperature will decrease because when I move the piston back, the distance apart each gas particle increases so the potential energy will increase (similar to boiling or melting when particle distance increases). So for internal energy to remain unchanged, the kinetic energy have to decrease explaining the decrease in temperature. So with the greater volume and lower temperature pressure has to be decreased.

You are right, both the pressure and temperature will decrease as the volume increases while the internal energy stays constant, although the change of temperature is very slight. The pressure would decrease even in case of no change of temperature: think of Boyle's Law. The reason is that the collisions of molecules with the walls are less frequent if the volume is greater. This collisions impart momentum to the wall, which causes the pressure.

ehild

 

[ehild]

I stand by what I said. The internal energy of a real gas is, in general, a function of both temperature and pressure, and deviates from a pure dependence on temperature under conditions where the gas deviates from PV = nRT. The dependence of internal energy on pressure can be precisely calculated from the gas' equation of state.

The internal energy of a gas is function of the chosen state variables (two at last). They can be entropy and volume, temperature and volume or temperature and pressure. As change of the temperature and pressure is the question in terms of change of volume, consider the energy as function of T and V. For ideal gas, U=CvT (Cv is the heat capacity at constant volume). For Van der Waals gas, U=CvT-aN²/V.

ehild

 

[Chestermiller]

The internal energy of a gas is function of the chosen state variables (two at last). They can be entropy and volume, temperature and volume or temperature and pressure. As change of the temperature and pressure is the question in terms of change of volume, consider the energy as function of T and V. For ideal gas, U=CvT (Cv is the heat capacity at constant volume). For Van der Waals gas, U=CvT-aN²/V.

ehild

In general, for an ideal gas U ≠ CvT, since Cv is a function of temperature. More correctly, dU = Cv dT. I haven't worked out what the equation is for dU for a Van der Waals gas, but I'm sure it is worked out in many textbooks.

You choose a reference state at low pressure (ideal gas region), and then you first calculate ΔU at constant pressure for the given ΔT. Then you calculate ΔU at constant T for the given ΔP. This gives you the overall ΔU of a specified state relative to the reference state. You can do this for any specified state, and then you can calculate changes in U from one specified state to another.