1.

2 moles of ideal monatomic gas is carried from a state (P_(0) , V_(0))to a state (2P_(0) , 2V_(0))along a straight line path in a P- Vdiagram. The amount of heat absorbed by

Answer»

<P>`3P_(0) V_(0)`
`9/2 P_(0) V_(0)`
`6P_(0) V_(0)`
`3/2 P_(0) V_(0)`

SOLUTION :
TOTAL work done forpath AB ,
`W_(AB) ` = are under curve`AB = P_(0)V_(0) + 1/2 P_(0) V_(0) = 3/2 P_(0)V_(0)`
Change in internal energy for PATH AB
`Delta U_(AB) = nC_(v) Delta T "" (. :. " for monatomic gas, " C_(V) = 3/2 R) `
` :.Delta U _(AB) = n 3/2 R ((4P_(0)V_(0))/(nR) - (P_(0)V_(0))/(nR)) `
`rArrDelta U_(AB) = (9P_(0)V_(0))/2`
` :. `The amount of heat absorbed by the gas in the process is ,
`Delta Q = W_(AB) + Delta U_(AB) = 3/2 P_(0)V_(0) + 9/2 P_(0)V_(0) = 6P_(0)V_(0)`


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