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Grade: 11
        One mole of a diatomic gas is heated under “ Kibolinsky Process” in which gas
pressure 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.Answer
the following question:
The molar specific heat of the gas?
6 years ago

Answers : (1)

Debojyoti Chaki
11 Points
							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.
						
3 years ago
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  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


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