Adiabatic Process: Temperature and Work

[DocDrakeRamoray]

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If a given process is adiabatic, would an increase in volume decrease temperature?

I thought it would (according to Kaplan, temp decreases because work is done to expand gas and gas loses internal energy, but at the same time the process is adiabatic---how does that work?), but EK Chemistry 1001 says since process is adiabatic (no energy is lost or added) internal kinetic energy stays the same and therefore temperature stays the same. I don't get it, how can an increase in volume not decrease the temp?


EK reasoning: Since Internal Energy= change in heat + work, and since no change in heat (adiabatic) then no change in internal enegy, NO WORK is done when process is adiabatic (according to the formula) Can someone explain how without work can expand gas?


I'm confused because I think EK says exactly opposite of what Kaplan says. 🙁

References:

Kaplan PS book page 253 ( answer to Question 7)

EK Chemistry 1001, I think questions 347 and 348.

 

[SBR249]

I think the temperature of the gas does decrease. Here's my reasoning

dE = dQ + dW

E = internal energy
Q = heat
W = work

if the process is adiabatic, then dQ = 0,

so dE = dW

According to my physical chemistry book when gas expands without losing or gaining heat, it's temperature decreases. We can think of the volume expansion and the concurrent temperature decrease as a hypothetical sequence of two steps:

1) constant volume decrease in temperature: because volume is constant dW1=0 for this step

followed by

2) isothermal decrease in volume: dW2 = C_v*dT where C_v is the constant volume specific heat

so overall, dE = dW = dW1+dW2 = CdT

The key things I would remember in the above explanation are:

1) dE = dQ + dW (where W is the work done by the surroundings on the system) --> first law of thermodynamics

2) In all cases where the volume changes (except compression/expansion against a vacuum), work is being done and hence will be nonzero.

 

[DocDrakeRamoray]

Thanks. So your answer is on the side of Kaplan. I thought so too before I came across EK explanation.

 

[SBR249]

Yes, I would agree with Kaplan. When I took the thermo semester of P Chem, I was taught that when there's a change in volume, unless the surroundings is vacuum, there's always work. Not sure why EK said no work is done in an adiabatic expansion as there is clearly a formula for it.

Edit: actually you can also take a look at wikipedia. Under the first figure (the graph), the caption reads:

"For a simple substance, during an adiabatic process in which the volume increases, the internal energy of the working substance must decrease"

http://en.wikipedia.org/wiki/Adiabatic_expansion

 

[DocDrakeRamoray]

Yes, I would agree with Kaplan. When I took the thermo semester of P Chem, I was taught that when there's a change in volume, unless the surroundings is vacuum, there's always work. Not sure why EK said no work is done in an adiabatic expansion as there is clearly a formula for it.

Edit: actually you can also take a look at wikipedia. Under the first figure (the graph), the caption reads:

"For a simple substance, during an adiabatic process in which the volume increases, the internal energy of the working substance must decrease"

http://en.wikipedia.org/wiki/Adiabatic_expansion

hey thanks for the link!

 

[DocDrakeRamoray]

Yes, I would agree with Kaplan. When I took the thermo semester of P Chem, I was taught that when there's a change in volume, unless the surroundings is vacuum, there's always work. Not sure why EK said no work is done in an adiabatic expansion as there is clearly a formula for it.

Edit: actually you can also take a look at wikipedia. Under the first figure (the graph), the caption reads:

"For a simple substance, during an adiabatic process in which the volume increases, the internal energy of the working substance must decrease"

http://en.wikipedia.org/wiki/Adiabatic_expansion

hey thanks for the link!

"For a simple substance, during an adiabatic process in which the volume increases, the internal energy of the working substance must decrease"

 

[chemtopper]

During an adiabatic change no heat flows in and out of the system.so total internal energy remains same.But it does not mean that Temperature remains constant .It changes as per the requirement.Like expansion causes cooling and compression causes heating.
In isothermal system we have temperature constant.

 

[BerkReviewTeach]

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If a given process is adiabatic, would an increase in volume decrease temperature?

...but EK Chemistry 1001 says since process is adiabatic (no energy is lost or added) internal kinetic energy stays the same and therefore temperature stays the same. I don't get it, how can an increase in volume not decrease the temp?


EK reasoning: Since Internal Energy= change in heat + work, and since no change in heat (adiabatic) then no change in internal enegy, NO WORK is done when process is adiabatic (according to the formula)

EK Chemistry 1001, I think questions 347 and 348.

They are definitely wrong in this example. A simple, everyday proof involves an aerosol cannister, which becomes cold after spraying, as the remaining gas expands to fill the container. While this is not exactly the same thing (given that material is lost from the closed system and there may be a phase change with some aerosol products), it emphasizes the basic idea behind the Carnot cycle. As particles expand (increase their intermolecuar distance, the intermolecular forces are weakened). Energy is necessary to overcome these forces (bond breaking is endothermic). If the system is adiabatic, then heat must be absorbed to account for the work energy needed to expand the gas.

I believe the EK author has mistakenly used adiabatic and isothermal as the same thing.

In general, errors like this can be so unnerving. Try not to let it impact your psyche too much.

 

[Will Hunting]

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If a given process is adiabatic, would an increase in volume decrease temperature?

I thought it would (according to Kaplan, temp decreases because work is done to expand gas and gas loses internal energy, but at the same time the process is adiabatic---how does that work?), but EK Chemistry 1001 says since process is adiabatic (no energy is lost or added) internal kinetic energy stays the same and therefore temperature stays the same. I don't get it, how can an increase in volume not decrease the temp?


EK reasoning: Since Internal Energy= change in heat + work, and since no change in heat (adiabatic) then no change in internal enegy, NO WORK is done when process is adiabatic (according to the formula) Can someone explain how without work can expand gas?


I'm confused because I think EK says exactly opposite of what Kaplan says. 🙁

References:

Kaplan PS book page 253 ( answer to Question 7)

EK Chemistry 1001, I think questions 347 and 348.

EK is right. The OP failed to mention they were talking about FREE ADIBIATIC expansion. In this case, there is NO change in internal energy because there is no Work. The gas is being released into a VACUUM which has 0 pressure. So, since work and heat are 0, Internal is the same.

However, in the case of reversible adibiatic expansion or compression, there is defintely a change in temperature. Yay for P-chem.