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ajprincess
 3 years ago
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
 3 years ago
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.

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1do you have a formula yet/

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1338339090808:dw

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1338339348203:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1and the question asks for \(\frac{\text d u}{\text dt}\)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0cant say id even know how to begin with this one :/ srry

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0it's ok. Thanxxx a lot anyways.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1it 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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1The 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 \[p=nk_bT\] \[p=\frac23u\] \(U=\frac 32nk_BT\) \(\Delta U=\frac 32nk_B\Delta T\)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1there are too many formulas in thermodynamics i found this subject very difficult

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0Bt y hav they given gamma @unklerhaukus.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i cant see how to use gamma for this question, the method of cooling has not be specified

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1338353654846:dw When I solve using this and the above formula i posted I end up with a different answer.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1what were the two answers you got?

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0one is 635,215.625 and the other is 2.30*10^20

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1and the units that belong to those numbers are ?

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0first one is Joules bt the second is the answer I found using ur formula. I am confused with its units.

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.0second answer has to be 2.55*10^22

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1in physics , units are most important , did you stick to SI units , and remember to convert temperatures to Kelvin

eashmore
 3 years ago
Best ResponseYou've already chosen the best response.0Let'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
 3 years ago
Best ResponseYou've already chosen the best response.1are you sure the pressure is constant, not the volume @eashmore?

eashmore
 3 years ago
Best ResponseYou've already chosen the best response.0The problem statement explicitly states "at a constant pressure of 1 atm." Counter intuitive? Sure, but I'll stick to the problem statement.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1ok that isa good point, but i still dont see how come you can assume reverseable cooling
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