A certain volume of a gas (diatomic) expands isothermally at `20^(@)C` until its volume is doubled and then adiabatically until its volume is again doubled. Find the final temperature of the gas, given `g = 1.4` and that there is `0.1 mole` of the gas. Also calculate the work done in the two coses `R = 8.3J mole^(-1) K^(-1)`

Answer

`1.4xx10^(2J)`

Explanation

We require T-V relation to calculate the final temperature.
We have`TV^(gamma-1)= constant//(273+20)V^(g-1)=(273+t)(2V)^(gamma-1)`
or `273+t=(293)/(2^(1.4-1))=(293)/(2)^(0.4)`
`log(273+t)=log293-0.4log2-0.4log2=log293-0.4xx0.3010`
`=2.4669-0.1204`
or `log(273+t)=2.3465`
or `273+t=antilog2.3465`
or `273+t=222.1`
`t=-50.9^(@)C`
(i) Work done in isothermal process
`=nRTlog_(e)(V_(2))/(v_(1))=(8.3)/(10)xx293log_(e)(2V)/(V) therefore(n=(1)/(10))`
`=0.83xx293xx2.3log_(10)^(2)(becauselog_(e)x=2.3log_(10)x)`
`=0.83xx293xx2.3xx0.3010=1.684xx10^(2)J`
(ii) Work done in adiabatic processs`= (nR(T-T'))/(gamma-1)=0.83(293-222.1)/(1.4-1)`
`=(0 .83xx70.9)/(0.4)=1.47xx10^(2)J`