The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process form an initial state `(P_1,V_1,T)` to the final state `(P_2,V_2 , T)` is equal to

Options

A. Zero
B. `R` `ln T`
C. `R" ln"(V_(1))/(V_(2))`
D. `R" ln"(V_(2))/(V_(1))`

Answer

D

Explanation

The Change in entropy of an ideal gas ,
`DeltaS=(DeltaQ)/(T)` ….(i)
In isothermal process , there is no change in internal energy of gas , I .e `DeltaU=0`
`therefore" " DeltaU=DeltaQ-W`
`rArr" " 0=DeltaQ-WrArrDeltaQ=W`
i.e `DeltaQ` = work done by gas in isothermal process which went through an isothermal process from `(p_(1),V_(1)T)` to `(p_(2),V_(2),T)`
`"or" " " Delta Q=muRT"log"_(e)((V_(2))/(V_(1)))` (ii)
For 1 mole of an ideal gas , `mu=1`
So, from Eqs. (i) and (ii) ,we get
`"or" " " DeltaS=R"log"_(e)((V_(2))/(V_(1)))=R "In"((V_(2))/(V_(1)))`