For silver, `C_(P)(J K^(-1)"mol"^(-1))=23+0.01T`. If the temperature `(T)` of `3` moles of silver is raised from `300K` to `1000K` at `1` atm pressure, the value of `DeltaH` will be close to :

Options

A. 21 kJ
B. 13 kJ
C. 16 kJ
D. 62 kJ

Answer

D

Explanation

`DeltaH = int nC_(p)dT = 3 int_(300)^(1000) (2.3 + 0.01 T) dT - 3[23T + (0.01T^(2))/2]_(30.0)^(1000)`
`=3[23000 + 0.01/2 (1000)^(2)-23(300)-(0.01(300)^(2))/2]`
`=3[23000 + 5000 - 6900 - 450] = 61950` J = 62 kJ