2 moles of a diatomic gas  are enclosed in a vessel. When a certain amount of heat is supplied,50% of gas molecules get dissociated, but there is not rise in temperature. What is the heat supplied, if temperature is T.a)RT     b)RT/2      3)11/2RT        4)5RT

arun
123 Points
7 years ago
as there is no rise in temprature so $\Delta U=0$
initial number of moles = 2
final number of moles = 3
now from first law of thermodynamics
$\Delta Q= \Delta U+P\Delta V$
$\Delta Q= \Delta U+\Delta n\times RT$
$\Delta Q= RT$                                       $\left [ \because \Delta n=1 \right ]$
so heat given is RT
Ashish
13 Points
4 years ago
Initially total internal energy Ui=no. of moles × 5R/2 (since it is a diatomic gas)
Finally after supplying heat there are 1 mole of diatomic gas and 2 moles of monoatomic gas.
So final internal energy is Uf=5R/2+2×3R/2=11R/2-10R/2 = R/2.

So heat supplied is 11R/2-10

Triple Fire
13 Points
4 years ago
Given;
Initial moles =2
Final moles = 3 , (1mole of diatomic and 2 moles of 50 % dissapiated monoatomic molecules.)
∆T =0 and P∆V=nR∆T
U=1/2nfRT ; f= degree of freedom.
∆Q=∆U-P∆V=>∆Q=∆U