1.

1 mole of anideal gas at25^(@) C is subjected to expand reversibly and adiabaticallyto ten times of its initialvolume. Calculate the change in entropy during expansion (in J K^(-1) mol^(-1))

Answer»

`19.15`
`-19.15`
`4.7`
zero

Solution :`DeltaS = nC_(v) l"N" (T_(2))/(T_(1)) + nRl"n" (V_(2))/(V_(1))`
For ADIABATIC process `(Q-0)`
`DeltaE = W "" IMPLIES""nC_(v) ln(T_(2))/(T_(1)) =-NR ln (V_(2))/(V_(1)) "" therefore""DeltaS=0`


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