The specific heats of iodine vapour and solid are `0.031` and `0.055 cal//g` respectively. If heat of sublimation of iodinde is `24cal//g` at `200^(@)C`, what is its value at `250^(@)C` ?

Options

A. `5.7calg^(-1)`
B. `11.4calg^(-1)`
C. `22.8calg^(-1)`
D. `45.6calg^(-1)`

Answer

C

Explanation

Solid `I_(2)` is converted into `I_(2)` vapour
`I_(2)(s)=I_(2)(g)DeltaH=24cal//g` at `200^(@)C`
`DeltaC_(P)=(C_(p))_(2)(g)=(C_(p))I_(2)(s)`
`=0.031-0.055=-0.24calg^(-1)`
From Kirchhoff's equation, we have
`DeltaH_(250^(@)C) -DeltaH_(200^(@)C) =DeltaC_(p)(T_(250^(@)C) - T_(200^(@)C))`
`DeltaH_(250^(@)C)=24-0.024(523-473)`
`=24-1.2=22.8calg^(-1)`