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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

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The correct Answer is:
D

d. `Delta Q = (K gamma)/(gamma - 1) = (KC_(P) // C_(V))/((C_(P)//C_(V)) - 1) = (KC_(P))/(C_(P) - C_(V))`
`= (K (n C_(P) Delta T))/((n C_(P) Delta T - nC_(v) Delta T)) = (K Delta Q)/((Delta Q - Delta U)) = (K Delta Q)/(Delta W)`
or `1 = (K)/(Delta w) implies K = Delta W`
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