Molar heat capacity of an ideal gas varies as `C = C_(v) +alphaT,C=C_(v)+betaV`
and `C = C_(v) + ap`, where `alpha,beta` and a are constant. For an ideal gas in terms of the variables `T` and `V`.

Options

A. `Ve^((alphaT//R))=const`
B. `T.e^((R//betaV))= const`
C. `V=anT`
D. `Va=nT`

Answer

A, B, C

Explanation

(i) From first law of temodynamics,
`dQ=dU+dW dQ=CdT=C_(v)dT+PdV`
`C=(dQ)/(dT)=C_(v)+P(dV)/(dT)=C_(v)+(RT)/(V)(dT)/(dT)…..(1)`
since it varie as, `C=C_(v)+propT ……(2)`
On comparing expression `(1) and (2)`.
we have `propT=((RT)/(V))((dV)/(dT))((prop)/(R))dT=(dV)/(V) .....(3)`
On integrating eqn. `(3)`, we have log `V-(propT)/(R)=` constant or V = (another constant) `e^(propT//R)`
or `V=Ve^-(propT//R)=` constant ......(4)
Eqn. `(4)` is the equation of process.
(ii) We have `C=C_(v)+(RT)/(V)(dV)(dT)`
Comparing it with given molar heat
capacity, `C=C_(v)+betaV` Hence, we have
`(RT)/(V)(dT)/(dT)=betaV(dV)/(V^(2))=(beta)/(R)(dT)/(T) ......(5)`
Above equation integration yields
`-(1)/(V)=(beta)/(R)logT+` constant
`logT+(R)/(betaV)=` another constant ....(6)
or `Te^((R//betaV))=` another cosntant ......(7)
(iii) On comparing `C=C_(v)+(RT)/(V)(dT)/(dT)`
and `C=C_(v)+aP` we have `(RT)/(V)(dT)/(dT)=aP , (RT)/(V)(dV)/(dT)=aP`
Since `PV=RT` for one mole, hence
`(dV)/(dT)=a or V =aT or V=an T` for n moles