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In an isobaric process, Delta Q = (K gam...

In an isobaric process, `Delta Q = (K gamma)/(gamma - 1)` where `gamma = C_(P)//C_(V)`. What is ` K`?

A

Pressure

B

Volume

C

`Delta U`

D

`Delta W `

Text Solution

Verified by Experts

The correct Answer is:
d
Given , `Delta Q = (Kgamma)/(gamma -1) = (KC_(P)//C_(V))/((C_(P)//C_(V)) -1) = (KC_(P))/(C_(P) - C_(V))`
`= (K(nC_(P)DeltaT))/((nC_(P)DeltaT - nC_(V)DeltaT)) = (KDeltaQ)/((DeltaQ - DeltaU)) = (KDelta Q)/(Delta W)`
or `1 = K/(DeltaW) rArr K = Delta W`
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