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2 moles of ideal monatomic gas is carrie...

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- V diagram. The amount of heat absorbed by

A

`3P_(0) V_(0)`

B

`9/2 P_(0) V_(0)`

C

`6P_(0) V_(0)`

D

`3/2 P_(0) V_(0)`

Text Solution

Verified by Experts

The correct Answer is:
c

Total work done for path 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)) `
`rArr Delta 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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