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 ofthe gas. Also calculate the work done in the two coses R = 8.3J mole^(-1) K^(-1)

A certain volume of a gas (diatomic) expands isothermally at 20^(@)C u

1.	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 ofthe gas. Also calculate the work done in the two coses R = 8.3J mole^(-1) K^(-1)
Answer» Solution :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`