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A gas is expanded form volume V(0) to 2V...

A gas is expanded form volume `V_(0) to 2V_(0)` under three different processes as shown in the figure . Process 1 is isobaric process process 2 is isothermal and and process 3 is adiabatic .
Let `DeltaU_(1),DeltaU_(2) and DeltaU_(3)` be the change in internal energy of the gs in these three processes then

A

`DeltaU_(1) gt DeltaU_(2) gt DeltaU_(3)`

B

`DeltaU_(1) lt DeltaU_(2) lt DeltaU_(3)`

C

`DeltaU_(2) lt DeltaU_(1) lt DeltaU_(3)`

D

`DeltaU_(2) lt DeltaU_(1) lt DeltaU_(3)`

Text Solution

Verified by Experts

The correct Answer is:
a
Process 1 : Isobaric , `Delta W = P Delta V " and " PV = n RT`
` :. P Delta V = n R Delta T , " so, " ( Delta W)/(n R) = Delta T`
So, `Delta U_(1) = (nR)/((gamma - 1)) * Delta T = (nR)/((gamma - 1)) (Delta W)/(nR)`
` rArr Delta U_(1) = (Delta W)/((gamma - 1))` ......(i)
Process 2 : Isothermal, `Delta T = 0, Delta U_(2) = 0 ` ...........(ii)
Process 3 : Adiabatic, `Delta Q = 0 , " so, " Delta W = - Delta U_(3)`
` :. Delta U_(3) = - Delta W = - (nR)/ ((gamma - 1)) Delta T` .... (iii)
` :. ` From (i) , (ii and (iii) we ger ,
`Delta U_(1) gt Delta U_(2) gt Delta U_(3)`
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