Plzz help.
Typical bedroom contains 2500 moles of air. Find the change in internal energy of the air, when it is cooled from 23.9 degree C to 11.6 degree C at a constant pressure of 1atm. Treat this air as an ideal gas with gamma = 1.4.

- ajprincess

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- UnkleRhaukus

do you have a formula yet/

- ajprincess

|dw:1338339090808:dw|

- ajprincess

|dw:1338339348203:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- UnkleRhaukus

and the question asks for \(\frac{\text d u}{\text dt}\)

- ajprincess

jst du.

- UnkleRhaukus

oh ok.

- amistre64

cant say id even know how to begin with this one :/ srry

- ajprincess

it's ok. Thanxxx a lot anyways.

- UnkleRhaukus

it is really annoying because it seams like a simple question, and i know i have done an almost identical question about a year ago, im gonna look through my old books

- ajprincess

ha k.

- ajprincess

thanxxx

- UnkleRhaukus

The pressure is constant
\(p\) is pressure; \(n=\frac NV\) is the moles per unit volume ;
\(k_b\)the Boltzmann constant; \(T\) is the Temperature in Kelvin;
\(u=\frac UV\) is the kinetic energy per unit volume
\[p=nk_bT\]
\[p=\frac23u\]
\(U=\frac 32nk_BT\)
\(\Delta U=\frac 32nk_B\Delta T\)

- ajprincess

Thanxxxxx a lottt

- UnkleRhaukus

there are too many formulas in thermodynamics
i found this subject very difficult

- ajprincess

:)

- ajprincess

Bt y hav they given gamma @unklerhaukus.

- UnkleRhaukus

i cant see how to use gamma for this question,
the method of cooling has not be specified

- ajprincess

|dw:1338353654846:dw|
When I solve using this and the above formula i posted I end up with a different answer.

- UnkleRhaukus

what were the two answers you got?

- ajprincess

one is 635,215.625
and the other is 2.30*10^-20

- UnkleRhaukus

and the units that belong to those numbers are ?

- ajprincess

first one is Joules bt the second is the answer I found using ur formula. I am confused with its units.

- ajprincess

second answer has to be 2.55*10^-22

- ajprincess

- UnkleRhaukus

in physics , units are most important , did you stick to SI units , and remember to convert temperatures to Kelvin

- ajprincess

ya I did.

- anonymous

Let's first make a couple of assumptions to better define the problem.
1) Closed system.
2) Constant specific heats
3) Ideal gas
4) Reversible
We also know that the process happens at constant pressure.
With Assumptions 1, 2, and 3, we can say that\[dU = m C_v dT\]However, we can look up the specific heat on a molar basis. The expression now becomes\[dU [ kJ] = n [mole] \cdot \bar C_v \left [ kJ \over {mol \cdot K } \right ] \cdot dT [K ~ or~ C]\] (Note that bar over the \(C_v\) indicates molar units.)
But, because the process happens at constant pressure and not constant volume, we cannot use the constant volume specific heat value (\(C_v\)). Let's recall that change in enthalpy can be expressed as (under conditions specified by Assumptions 1, 2, and 3) \[dH = m C_p dT = n \bar C_p dT\]Let's also recall that enthalpy can be expressed in terms of internal energy as\[H = U + pV\]Differentiating, we obtain \[dH = dU + dp \cdot V + p \cdot dV\]Since pressure is constant, \(dp \cdot V = 0\)\[dH = dU + p \cdot dV\]Therefore, \[dU = n \bar C_p dT - p \cdot dV\]From Assumption 3, we can express the change in volume as\[dV = {nR(dT) \over p}\]\[\therefore dU = n \bar C_p (dT) - {nR(dT) \over p}\]With Assumption 2 and 3 and the fact that pressure and temperature are state quantities, we can expression the differential as\[\Delta U = n \bar C_p (\Delta T) - {n R (\Delta T) \over p}\]
We don't know \(C_p\) however, but we know gamma and the gas constant. Let's work a couple of relations to get \(C_p\) in terms of gamma and R. It can be shown that\[C_p = {\gamma n R \over \gamma - 1}\]
We obtain the following final expression\[\Delta U = n \left [ \gamma n R \over \gamma -1 \right] \Delta T - {n R \Delta T \over p}\]
Be sure the units on R matches the units on n and p.

- UnkleRhaukus

are you sure the pressure is constant, not the volume @eashmore?

- anonymous

The problem statement explicitly states "at a constant pressure of 1 atm." Counter intuitive? Sure, but I'll stick to the problem statement.

- UnkleRhaukus

ok that isa good point, but i still dont see how come you can assume reverseable cooling

- ajprincess

Thanxxxx a lottt.

Looking for something else?

Not the answer you are looking for? Search for more explanations.