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One mole of an ideal monoatomatic gas is...

One mole of an ideal monoatomatic gas is taken round the cylic process `ABCA` as shown Then

A

Work done by the gas `P_(0)V_(0)`

B

Heat rejected in the process `CA` is `(5P_(0)V_(0))/(2)` and absorbed in the process `AB` is `3P_(0)V_(0)`

C

Heat absorbed in the process `BC` is `(P_(0)V_(0))/(2)`

D

`(25)/(8)(P_(0)V_(0))/(R)` is the maximu temperature attained by the gas during the cycle

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(a). The work done by the gas is equal to the area inder the closed curve. Thus work done in cycle is `W=P_(0)V_(0)`
(b) Heat rejected in path `CA` is given as
`Q_(CA)=nC_(P)DeltaT=(5)/(2)P_(0)V_(0)`
Heat absored in path `AC` is
`Q_(AC)=nC_(P)(T_(B)-T_(A))=3P_(0)V_(0)`
(c) For cycle `ABC`, we have heat supplied = work done
by the gas `-((5)/(2))P_(0)V_(0)+3P_(0)V_(0)+Q_(BC)=P_(0)V_(0)`
`implies Q_(BC)=P_(0)V_(0)//2` .
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