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The volume of 1 mol of an ideal gas with...

The volume of 1 mol of an ideal gas with the adiabatic exponent `gamma` is changed according to the relation `V = b/T` where b = constant. The amount of heat absorbed by the gas in the process if the temperature is increased by `Delta T` will be

A

`((1-gamma)/(gamma +1)) R Delta T`

B

`R/(gamma - 1) Delta T`

C

`((2-gamma)/(gamma - 1)) R Delta T`

D

`(R Delta T)/(gamma - 1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`V = b/T`
`:. dV = -b/(T^2) dT`
Work done , `W = int PdV = -int (RT)/(V) cdot b/(T^2) dT`
` = -int (RT)/V cdot (VT)/(T^2) dT = - R int dT = -R delta T`
Change in internal energy,
`Delta U = int C_(v)dT = R/(gamma - 1) int dT = R/(gamma - 1) Delta T`
`["As" R/(gamma - 1) = (C_(p) - C_(v))/((C_p)/(C_v) -1) = (C_(p)-C_(v))/((C_(p)-C_(v))/(C_v)) = C_(v)]`
`:. "Heat absorbed" , Q = Delta U + W = (R/(gamma - 1)-R) Delta T`
`= R xx (1-gamma + 1)/(gamma - 1) Delta T = ((2-gamma)/(gamma - 1)) R Delta T`.
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