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One mole of a diatomic gas is heated under “ Kibolinsky Process” in which gaspressure is temperture controlled according to law P=4/5*CT^(3/2) . Where C is Kibolinsky constant. In this process, gas is heated by 300 K temperature.Answerthe following question:The molar specific heat of the gas?
Answer is 2RHow?Sol.P=4/5CT^3/2......(1)(PV)^3/2=(nRT)^3/2.......(2)(1)/(2)= P(V^3/2)=const........n=3C=R/gama-1+R/1-nPutting the values we get C=2R.
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