Change in Velocity does what to temperature?

[destroythemcat]

A question asked: If we decrease the velocity by 2 of a gas in a closed container what happens to the Temperature?

A: In the explanation the book explained that we cannot use PV=nRT bc we cannot assume Pressure is constant. Instead we have to use U=q+w, and we cancel q --> change in U=w --> change in U=3/2nRT.
So Temperature goes up because work is being done on the system.

logically it makes sense. Work is done=internal energy goes up=temp goes up.

But why can we use the equation U=w When Work=P x change in V -->we also must assume pressure is not changing?!!?

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[JMMTB]

Do you mean velocity or Volume?

If you decrease the velocity of the gas molecules their Kinetic Energy would be decreased and the Temperature as well.
If you decrease the volume that could decrease the Temperature depending on the circumstances.
PV=nRT uses pressure, volume, moles of gas, the gas constant, and Temperature (in Kelvins)

The change in heat for constant pressure is dQ=nCpdT where n=moles of gas. Cp=Molar specific heat of gas when constant P which equals (5/2)R for monatomic (He, Ne,...) or 7/2R for diatomic (N2, O2, H2...)

For change in heat with constant volume dQ=nCvdT where Cv is (3/2)R for monatomic and (5/2)R for diatomic

Total dE=nCvdT always. Always.

So if Volume is constant all Q that goes into or out of the system is also equal to all of the E that goes into or out of the system. With no change in the Volume there is no work being done on the gas or done by the gas.

If we have constant P then E=nCvdT and Q=nCpdT. To find work then E=Q+W and W=E-Q so dW=n(Cv-Cp)dT=-nRdT

If it is an adiabatic process (no heat flow dQ=0) then dE=W=nCvdT

And lastly if the Temperature is constant dE=0 and Q=W

Hope that helps!

 

[destroythemcat]

You seem to really get this….I meant That volume decreases sorry! So what equations would i use then?

 

[Mad Jack]

Do you mean velocity or Volume?

If you decrease the velocity of the gas molecules their Kinetic Energy would be decreased and the Temperature as well.
If you decrease the volume that could decrease the Temperature depending on the circumstances.
PV=nRT uses pressure, volume, moles of gas, the gas constant, and Temperature (in Kelvins)

The change in heat for constant pressure is dQ=nCpdT where n=moles of gas. Cp=Molar specific heat of gas when constant P which equals (5/2)R for monatomic (He, Ne,...) or 7/2R for diatomic (N2, O2, H2...)

For change in heat with constant volume dQ=nCvdT where Cv is (3/2)R for monatomic and (5/2)R for diatomic

Total dE=nCvdT always. Always.

So if Volume is constant all Q that goes into or out of the system is also equal to all of the E that goes into or out of the system. With no change in the Volume there is no work being done on the gas or done by the gas.

If we have constant P then E=nCvdT and Q=nCpdT. To find work then E=Q+W and W=E-Q so dW=n(Cv-Cp)dT=-nRdT

If it is an adiabatic process (no heat flow dQ=0) then dE=W=nCvdT

And lastly if the Temperature is constant dE=0 and Q=W

Hope that helps!

Oh thank god, I was like, less velocity will equal less temp, not more, unless I'm losing my mind.

 

[JMMTB]

Is it adiabatic (no heat flow)? Because you said that the solution says to ignore q.

Lets say it is adiabatic. We would actually use some more Equations if you wanted to know the exact dT

PiVi^(gamma)=PfVf^(gamma)

Pi=Pressure initial and f=final...
gamma=Cp/Cv
Gamma (monatomic)=5/3
Diatomic=7/5

You could also use
TV^(gamma-1)=TV^(gamma-1)

Although I'm like 99% sure the MCAT would never have you make those type of calculations.

In that case remember like you said if you are doing work on the system, it gains energy. T will go up.

Kind of a, well, interesting question because say when you compress the gas, all of the energy that is added by W is expelled as heat to its surroundings then T would be constant. It should have a tad bit more information.