Thermodynamics; find the thermal energy

[iScience]

question:Calculate the total thermal energy in a liter of helium at room temperature and atmospheric pressure. Then repeat the calculation for a liter of air.



I'm just confused because i thought thermal energy only depended on the translational kinetic energy of the particles. So why do i need all the pressure if the temperature is already given?

the only equation that comes to mind is E(kinetic)=3/2kT

and ...maybe the gas law?..

where do i go from here?

 

[janhaa]

Maybe this way:

U(thermal) = N*1,5kT

and the ideal gas law: pV = NkT

 

[iScience]

so E(thermal)=(3/2)NkT -----> E(thermal)=(3/2)PV? since N=PV/kT? and then i just plug and chug?

 

[janhaa]

so E(thermal)=(3/2)NkT -----> E(thermal)=(3/2)PV? since N=PV/kT? and then i just plug and chug?

Yes, I believe so...

 

[iScience]

isn't a "PV" term dynamically associated with the pressure with respect to a change in volume? ie calculating the work done on a system from a PV diagram? (ie the area under the PV curve)

 

[DrClaude]

Maybe this way:

U(thermal) = N*1,5kT

and the ideal gas law: pV = NkT

You're missing something here. The correct formula for an ideal gas is
$$
U = \frac{f}{2} N k T
$$
where $f$ is the number of (quadratic) degrees of freedom. That is why you get a different answer for helium and air.

 

[DrClaude]

isn't a "PV" term dynamically associated with the pressure with respect to a change in volume? ie calculating the work done on a system from a PV diagram? (ie the area under the PV curve)

Yes, expansion/contraction work done by/on a gas is obtained from
$$
W = - \int_{V_i}^{V_f} P \, dV
$$
but $PV$ by itself is just the product of the pressure and the volume.

 

[iScience]

but $PV$ by itself is just the product of the pressure and the volume.

well what i was getting at was i thought that that quantity (PV) was the case where the P is constant (isobaric) but still a dynamic case where the Volume is changing. so basically i don't understand why the quantity PV is used for a static case.

 

[DrClaude]

well what i was getting at was i thought that that quantity (PV) was the case where the P is constant (isobaric) but still a dynamic case where the Volume is changing.

Pressure doesn't have to be constant. The formula for work is valid even when $P$ varies, although this might make it complicated to calculate the integral (unless $P$ can be expressed as a simple function of $V$).

so basically i don't understand why the quantity PV is used for a static case.

Equations of state are equations that relate the different macroscopic observables of a system. In the case of a gas, these observables are $P$, $V$, and $T$ (for a fixed quantity of gas). For an ideal gas, the relation is exactly
$$
PV = N k T
$$
or
$$
PV = n R T
$$
Such equations of state also exist for more realistic gases: they are slightly more complicated, but again relate $P$, $V$, and $T$, such that if you fix two of them you can know the value of the third.

As an example, if you measure the pressure inside a bicycle tire and know what the temperature is, then you can calculate the volume inside the inner tube. So you see, this has nothing to do with "dynamics."

 

[Chestermiller]

This is the first response to this thread in over a year and a half. I am closing this thread.

Chet