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An ideal gas (1 mol, monatomic) is in th...

An ideal gas `(1` mol, monatomic) is in the intial state `P` (see Fig.) on an isothermal `A` at temperature `T_(0)`. It is brought under a constant volume `(2V_(0))` process to `Q` which lies on an adiabatic `B` intersecting the isothermal `A` at `(P_(0),V_(0),T_(0))`. The change in the internal energy of the gas during the process is (in terms of `T_(0)) (2^(2//3) = 1.587)`

A

`2.3T_(0)`

B

`-4.6T_(0)`

C

`-2.3T_(0)`

D

`4.6T_(0)`

Text Solution

Verified by Experts

The correct Answer is:
b
Temperature at state `P = T_(0)`,since P lies on the isothermal of temperature `T_(0)` . If I be the temperature at Q. then for the adiabatic process B, we have
`T_(0) V_(0)^(gamma - 1)`
`rArr T = T_(0)/(2^(gamma -1)) = T_(0)/2^(2//3) [:. gamma = 5/3, " so " gamma - 1 = 2/3]`
Change in the internal energy of the gas is
`Delta U = V_(V) (T - T_(0)) = (R/(gamma -1)) (T_(0)/2^(2//3) - T_(0))`
` = (3RT_(0)(1 -2^(2//3)))/(2 xx 1.587) = - 0.55 RT_(0) = - 4.6T_(0)`
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