isothermal expansion of a gas

[mcgill2012]

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just want to make sure im thinking of this properly.

delta U = Q - W

since its isothermal, theres no temperature change and therefore deltaU = 0.

Q = W

work done is +ve since work is being done by the gas. why is Q positive? because heat is flowing into the system in order to drive the work (expansion)?

 

[Rabolisk]

Everything you have said is right. For an ideal gas, U is directly proportional to temperature.

 

[mcgill2012]

thank you! in terms of ideal gases i know that under isobaric conditions, temperature increase is directly proportional to volume increase. im having trouble understanding this conceptually though - why would an increase in volume coincide with an increase in temperature? this is counterintuitive to me.

 

[Rabolisk]

That comes from the Ideal Gas Law, PV = nRT. Regardless of how you get there, if you increase volume at constant pressure, you must increase temperature by the same factor. Relating this to the 1st law of thermodynamics, internal energy also increases by the same factor. You also know that work was done by the gas, and thus heat was added into the system. In fact, you know that more heat was added than work was done. However, you can't state the value(s) of heat and work, because they are not state functions, only the sum of heat and work (internal energy) is.

To sum it up, isobaric expansion of ideal gases are neither isothermal nor adiabatic.

 

[mcgill2012]

thank you!

 

[sangria1986]

thank you!

do we have to know isothermal and isobaric...its not mentioned in the chem topics of the AAMC....

 

[sangria1986]

what is adiabatic?

Also another way to think of isobaric is this: Since pressure is constant and volume is increasing how can this be? We know that in a normal process an increase in volume would necessarily decrease in pressure...but since pressure is constant, this means that particles within the volume (since it has expanded) are interacting with one another more since the number of particles hitting the container (measure of pressure) has not changed. Particles interacting with another at an increased level means greater temperature (more kinetic energy). This too is of course only applicable to ideal gasses.

 

[Rabolisk]

I am pretty sure you have to know what those terms mean. At the very least, you should be able to use them for thermodynamic calculations, if the passage or the question tells you what they mean.

I am not sure that I would characterize the particles as "interacting" more with one another at higher temperature. The kinetic energy of the particles increase, increasing the average velocity of the particles. This means more frequent collisions, as well as collisions with greater momentum, resulting in increased pressure.

Adiabatic means no heat exchange between system and surrounds. q = 0

 

[sangria1986]

I am pretty sure you have to know what those terms mean. At the very least, you should be able to use them for thermodynamic calculations, if the passage or the question tells you what they mean.

I am not sure that I would characterize the particles as "interacting" more with one another at higher temperature. The kinetic energy of the particles increase, increasing the average velocity of the particles. This means more frequent collisions, as well as collisions with greater momentum, resulting in increased pressure.

Adiabatic means no heat exchange between system and surrounds. q = 0

sorry i should have said collisions instead of interacting. I thought pressure was only due to particles colliding with the container-at least thats the basis for the ideal gas behavior correct? This is why ideal gas laws work best at high temperatures-since the collisions at high temperatures happen often enough with the container this becomes more true.

 

[Rabolisk]

Right. When I said more frequent and higher energy collisions, I meant collisions with the walls.