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Thermal Physics

Blazing goIITian

Joined:	29 Mar 2008
Post:	1024

14 Oct 2008 00:08:22 IST

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892

Heat

None

The volume of one mole of an ideal gas with adiabatic exponent is varied according to the expression where a is a constant.Find the amount of heat obtained by the gas in this process if the gas temperature is raised by temperature

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Comments (5)

aNdRoMeDa

Blazing goIITian

Joined: 10 Sep 2007

Posts: 1319

14 Oct 2008 00:28:08 IST

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tukka!

VT=constnt bt for adiabatic process TV^(gamma-1)=constnt

thus gamma -1=1............. thus gamma=2....

nw delQ=delE+w

w=nRdelT/(gamma-1).....................n=1

and delE=nCvdelT

thus delQ=CvdelT+R delT

thus delQ=delT(Cv+R)

bt meyers eqaution u get

delQ=delT*Cp

nw gamma=Cp/Cv ......................Cp=Cv+R

thus 2Cv=Cv+R thus Cv=R

and Cp=2R

thus delQ=delT*(2R)

correct me if m wrng

anchit saini

Blazing goIITian

Joined: 1 Feb 2008

Posts: 1251

14 Oct 2008 13:04:17 IST

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$TV=a\\ \\ -> PV^2 = aR = const \\ \\ ->x= 2 \\ \\ ->C_v = \frac{R}{\gamma - 1} - \frac{R}{x-1} \\ \\ = \frac{R}{\gamma - 1} - R\\ \\ =R\frac{2 - \gamma}{\gamma - 1} \\ \\ dU= C_v dT = R\frac{2 - \gamma}{\gamma - 1}dT\\ \\ dW=PdV = \frac{RT^2}{a}* \frac{-a}{T^2}dT =-RdT \\ \\ dQ=dU +dW = R\frac{2 - \gamma}{\gamma - 1}dT -RdT \\ \\ = R\frac{3 - 2\gamma}{\gamma - 1}dT$

PS->isn't gamma given , cos otherwise all the above seems to be fruitless

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

14 Oct 2008 15:18:05 IST

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adiabatic exponent varied ..explain wat is this..??

@ isnt tht polytropic's C ..?? and not Cv..

anchit saini

Blazing goIITian

Joined: 1 Feb 2008

Posts: 1251

14 Oct 2008 15:23:58 IST

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lol , seems i was sleeping !!
otherwise the answer is simply Cdt , and C has been calculated

reddevil_2009

Blazing goIITian

Joined: 29 Mar 2008

Posts: 1024

19 Oct 2008 23:07:07 IST

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answer is

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