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Asked by Prashant DIGHE | 01 Apr, 2020, 22:48: PM
answered-by-expert Expert Answer
Process Equation :-  P5 / T2 = C  , where C is constant, Hence P5 = CT2 ......................(1)
 
From first law of thermodynamics, we have, dQ = CvdT + P dV  ......................(2)
 
For ideal gas of 1 mole, we have,  P V = R T ......................(3)
 
By differentiating eqn.(3) , P dV + V dP = R dT .................(4)
 
We rewrite Eqn.(2) by Substituting PdV from eqn.(4)
 
dQ = CV dT + R dT - V dP .............................(5)
 
if we differentiate process eqn.(1), we get,  5 P4 dP = (2C) T dT
 
Hence from above eqn., we get,  V dP = (2C / 5 ) P-4 V  T dT = (2C/5) P-5 PV T dT ....................(6)
 
By substituting PV = RT   and P5 = CT2 using eqn.(3) and (1),  above eqn.(6) is simplified as, 
 
V dP = (2/5) R dT   ............................(7)
 
By substituting V dP using eqn.(7), we rewrite eqn.(5) as ,  dQ = CVdT + R dT - (2/5)R dT  ....................(8)
 
for monoatomic gas, CV = (3/2)R
 
Hence Eqn.(8) becomes,  dQ = [ (3/2)R + R - (2/5) R ] dT = (21/10)R dT
 
Hence molar specific heat = (21/10)R
Answered by Thiyagarajan K | 02 Apr, 2020, 07:18: AM

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