'm' grams of a gas of a molecular weight M is flowing in an isolated tube with velocity V. If the flow of the gas is suddenly stopped the rise in temperature is ( Gamma=ratio of specific heats, R = universal gas const, J =mechanical equivalent of heat): 1) MV^2(Gamma-1)/ 2RJ , 2) m/M V^2(gamma-1)/2RJ , 3)mV^2 gamma/2RJ, 4) MV^2gamma/2RJ My efforts: Work = J Q or 1/2mV^2 = J (m s x rise in temp) or Rise in temp = mV^2 /2m s Now how to convert S =specific heat into ratio of specific heats of gas ? we know gamma= (Cp /Cv)=(cp /cv) but not able crack further.

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'm' grams of a gas of a molecular weight M is flowing in an isolated tube with velocity V. If the flow of the gas is suddenly stopped the rise in temperature is ( Gamma=ratio of specific heats, R = universal gas const, J =mechanical equivalent of heat): 1) MV^2(Gamma-1)/ 2RJ , 2) m/M V^2(gamma-1)/2RJ , 3)mV^2 gamma/2RJ, 4) MV^2gamma/2RJ My efforts: Work = J Q or 1/2mV^2 = J (m s x rise in temp) or Rise in temp = mV^2 /2m s Now how to convert S =specific heat into ratio of specific heats of gas ? we know gamma= (Cp /Cv)=(cp /cv) but not able crack further.

Physics
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You may try to use Cp-Cv=R along with the ratio equation. For specific heats Sp=Cp/M and Sv=Cv/M
Cp-Cv=R or Cp/Cv-1=R/Cv or Cv= R/(Lamda-1) or Sv = R/M(Lamda-1) and the question is solved. But my doubt is why Cv is to be taken ? Is there a relation with 'isolated tube' of 'suddenly stopped'? Pl help
As you already know the kinetic energy of the moving gas is converted into internal energy ie heat of the gas when it is shut off. Since we have a constant volume we know \[mc _{v}\Delta T=\Delta E\] we also know that\[c _{p}=c _{v}+R/M\] and\[\gamma =c _{p}/c _{v}\] eliminate Cv in favor of gamma

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Thanks both Diwakar & gleem.

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