ajprincess
  • ajprincess
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UnkleRhaukus
  • UnkleRhaukus
do you have a formula yet/
ajprincess
  • ajprincess
|dw:1338339090808:dw|
ajprincess
  • ajprincess
|dw:1338339348203:dw|

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More answers

UnkleRhaukus
  • UnkleRhaukus
and the question asks for \(\frac{\text d u}{\text dt}\)
ajprincess
  • ajprincess
jst du.
UnkleRhaukus
  • UnkleRhaukus
oh ok.
amistre64
  • amistre64
cant say id even know how to begin with this one :/ srry
ajprincess
  • ajprincess
it's ok. Thanxxx a lot anyways.
UnkleRhaukus
  • 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
  • ajprincess
ha k.
ajprincess
  • ajprincess
thanxxx
UnkleRhaukus
  • 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
  • ajprincess
Thanxxxxx a lottt
UnkleRhaukus
  • UnkleRhaukus
there are too many formulas in thermodynamics i found this subject very difficult
ajprincess
  • ajprincess
:)
ajprincess
  • ajprincess
Bt y hav they given gamma @unklerhaukus.
UnkleRhaukus
  • UnkleRhaukus
i cant see how to use gamma for this question, the method of cooling has not be specified
ajprincess
  • ajprincess
|dw:1338353654846:dw| When I solve using this and the above formula i posted I end up with a different answer.
UnkleRhaukus
  • UnkleRhaukus
what were the two answers you got?
ajprincess
  • ajprincess
one is 635,215.625 and the other is 2.30*10^-20
UnkleRhaukus
  • UnkleRhaukus
and the units that belong to those numbers are ?
ajprincess
  • ajprincess
first one is Joules bt the second is the answer I found using ur formula. I am confused with its units.
ajprincess
  • ajprincess
second answer has to be 2.55*10^-22
ajprincess
  • ajprincess
@unklerhaukus
UnkleRhaukus
  • UnkleRhaukus
in physics , units are most important , did you stick to SI units , and remember to convert temperatures to Kelvin
ajprincess
  • ajprincess
ya I did.
anonymous
  • 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
  • UnkleRhaukus
are you sure the pressure is constant, not the volume @eashmore?
anonymous
  • 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
  • UnkleRhaukus
ok that isa good point, but i still dont see how come you can assume reverseable cooling
ajprincess
  • ajprincess
Thanxxxx a lottt.

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