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haridas_mandal

  • 3 years ago

'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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  1. Diwakar
    • 3 years ago
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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

  2. haridas_mandal
    • 3 years ago
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    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

  3. gleem
    • 3 years ago
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    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

  4. haridas_mandal
    • 3 years ago
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    Thanks both Diwakar & gleem.

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