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foxi

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Lets say we have a bioreactor (closed system) with cells floating in it. If we are to introduce a stir rod, which would increase the velocity of the cells within the system, would the pressure of the system increase as well?

I believe the pressure will increase: the particles are moving faster -> this means that KE goes up (since KE is directly related to velocity) -> this means that T is higher -> this means that P is higher (since volume remained constant)!

Right or wrong?
 

FutureD0C17

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Sounds correct to me. You are adding KE by stirring, which increases the temp and since the system is closed, the only component that can increase is pressure. pv = nrt
 

aldol16

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Increasing the temperature via stirring would increase the pressure, but the question is whether that effect would be measurable at all. If it's measurable, it's certainly not significant, otherwise we would encounter real problems when trying to stir any reaction at room temperature. So at the molecular level, sure, that would raise the kinetic energy of the particles. The amount of kinetic energy it adds though is tiny. Molecules are moving around pretty quickly even in fluids and have a distribution of speeds. So adding a little velocity just by stirring is not likely to significantly increase temperature. That's why we don't worry about stir bars going in our reactions, even when the reaction is heat-sensitive and performed at -78.
 

FutureD0C17

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Increasing the temperature via stirring would increase the pressure, but the question is whether that effect would be measurable at all. If it's measurable, it's certainly not significant, otherwise we would encounter real problems when trying to stir any reaction at room temperature. So at the molecular level, sure, that would raise the kinetic energy of the particles. The amount of kinetic energy it adds though is tiny. Molecules are moving around pretty quickly even in fluids and have a distribution of speeds. So adding a little velocity just by stirring is not likely to significantly increase temperature. That's why we don't worry about stir bars going in our reactions, even when the reaction is heat-sensitive and performed at -78.

What exactly is a closed system? Are we talking about a system that can only transfer heat and not matter, or not transfer either? Originally I was thinking transfer was totally excluded, however, that may instead be defined as an isolated system.

If heat can transfer, then the minimal amount generated by stirring should easily transfer away from the system, likely holding P constant. Your example of a stir bar doesn't really fit the bill since both forms can transfer.
 
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aldol16

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What exactly is a closed system? Are we talking about a system that can only transfer heat and not matter, or not transfer either? Originally I was thinking transfer was totally excluded, however, that may instead be defined as an isolated system.

If heat can transfer, then the minimal amount generated by stirring should easily transfer away from the system, likely holding P constant. Your example of a stir bar doesn't really fit the bill since both forms can transfer.

By definition, a closed system can exchange heat but not matter with the surroundings.

P will remain measurably constant throughout because the temperature gain will be negligible. Whether it radiates away depends on a variety of factors, including the thermal conductivity of the material.
 

NextStepTutor_2

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Hi @FutureDoc17

In classical mechanics, a closed system is a physical system that doesn't exchange any matter with its surroundings, and isn't subject to any force whose source is external to the system (more common in the Newtonian mechanics Qs on the MCAT).

A closed system in classical mechanics would be considered an isolated system in thermodynamics. This means mass is conserved within the boundaries of the system, but energy is allowed to freely enter or exit the system.

Technically, you might increase P a negligible amount, but for the purposes of the MCAT, the best answer to this question would be no, there is no real change in pressure.

Hope this helps, good luck!
 

aldol16

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A closed system in classical mechanics would be considered an isolated system in thermodynamics. This means mass is conserved within the boundaries of the system, but energy is allowed to freely enter or exit the system.

A thermodynamic isolated system exchanges neither mass nor energy with its surroundings... It looks like you switched the definition of a thermodynamic closed system with that of a thermodynamic isolated system. Above, you also state that a classical mechanical closed system is not subject to forces external to the system - that means energy cannot enter to leave the system, by the work-energy theorem.
 
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NextStepTutor_2

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A thermodynamic isolated system exchanges neither mass nor energy with its surroundings... Perhaps you switched up the terms?
Was it confusing? My apologies. Perhaps because I switched from classical mechanics to thermodynamics/chemistry and point may have been lost. Many students confuse these terms and their exact definitions between thermodynamics, physics, and classical mechanics.

For the purposes of the MCAT, one will most likely only need the thermodynamics definitions:

If a system is isolated, then nothing can enter or leave. Its energy and matter remain the same (IIRC, this is similar to a "closed" system in classical mechanics, someone else can elaborate). Any changes go on inside the system, and since it is isolated, we cannot know anything about an isolated system from the outside.

We typically think of the earth as a system or we might focus on a chemical reaction. Usually energy and matter can change in a system. The human body is a complex biological system which interacts with the surroundings, or think of the burning of mass in a bomb calorimeter. Systems can be complicated or simple. A bomb calorimeter only allows heat to be exchanged. Such a system is called closed. Another example of a closed system is a balloon being heated so the gas inside expands it, or a piston like in a car engine, no matter is exchange, only heat.

hope this helps, good luck!
 
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aldol16

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Was it confusing? My apologies. Perhaps because I switched from classical mechanics to thermodynamics/chemistry and point may have been lost. Many students confuse these terms and their exact definitions between thermodynamics, physics, and classical mechanics.

Yeah, this is great - your first post was just incorrect in that it stated: "This means mass is conserved within the boundaries of the system, but energy is allowed to freely enter or exit the system" when you described an isolated system. This is incorrect because an isolated system conserves mass and energy. You correct it in your second post, when you say: "If a system is isolated, then nothing can enter or leave. Its energy and matter remain the same."
 

NextStepTutor_2

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Yeah, this is great - your first post was just incorrect in that it stated: "This means mass is conserved within the boundaries of the system, but energy is allowed to freely enter or exit the system" when you described an isolated system. This is incorrect because an isolated system conserves mass and energy. You correct it in your second post, when you say: "If a system is isolated, then nothing can enter or leave. Its energy and matter remain the same."
Thanks for the assist. Lifelong learning FTW!
 
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