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Enthalpy of sublimation of iodine is $40 c a l g^{- 1} a t 200^{0} C$ . If specific heat of $I_{2} (s) a n d I_{2} (v a p)$ are 0.055 and 0.031 cal $g^{- 1} K^{- 1}$ respectively, then enthalpy of sublimation of iodine at $250^{0} C$ in cal $g^{- 1}$ is ______ [nearest integer]

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answer is 39.

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Detailed Solution

$Δ C_{p} = \frac{H_{2} - H_{2}}{T_{2} - T_{1}} I_{2 (s)} \to I_{2 (g)}, Δ I_{1} = 40 c a l / g a 200^{0} C Δ H_{2} = Δ H_{1} + C_{P_{r}} (T_{2} - T_{1}) = 40 + (0.031 - 0.055) \times (250 - 200) = 40 - 1.2 = 38.8 c a l / g r a m$

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