• anonymous
'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
• Stacey Warren - Expert brainly.com
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SOLVED
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