For a perfectly crystalline solid `C_("p,m") =T^(3)+bT`, where a & b are constant . If `C_("p,m")` is `0.04 J//k-"mole"` at 10 K and io `0.92 J//K-"mole"` at 20 k, Then molar entropy at 20 k is.

Options

A. `0. 92 J//K -mol`
B. `8.66 J// K-mol`
C. `0.813 J//K-mol`
D. None

Answer

C

Explanation

`.04 = aT_(1)^(3)+ bT_(1) " "rArr " "0.40 = a xx (1000) + b xx 10`
`" "rArr " "0.4 = 1000a + 10b" ".......(i)`
`0.92 =aT_(2)^(3) + bT_(2) " " rArr " "0.92 = a xx 8000 + 20b" "......(ii)`
On solving
`0.12 = 6000 a " "," "a= xx 10^(-5)`
`0.40= 2 xx 10^(-5) xx 1000+b+10`
`therefore b = 0.038`
`S_(m)= int (aT^(3) + bT)/(T)dT " "rArr " " (a[T_(2)^(3)- T_(1)^(3)])/(3) + b[T_(2) - T_(1)]`
`" "rArr (2xx 10 ^(-5) xx (8000-0))/(3) +b(20) rArr (2xx 10^(-5) xx 8000)/(3) +0.038xx(20)`
`" "0.053 + 0.76 rArr 0.813 J//K-mol`