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The specific heats of iodine vapour and solid are 0.031 and 0.055 cal / g respectively. If heat of sublimation of iodine is 24 cal/ g at 200 $°$ C , what is its value at 250 $°$ C ?

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2.28 cal / g

22.8 cal / mol

0.228 cal / g

22.8 cal / g

answer is C.

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Detailed Solution

$∆ C_{p} = \frac{∆ H_{T_{2}} - ∆ H_{T_{1}}}{T_{2} - T_{1}}$

$∆ C_{P} = - 0.055 + 0.031 = - 0.024$

$\Rightarrow - 0.024 = \frac{∆ H_{T_{2}} - 24}{250 - 200}$

$\Rightarrow - 0.024 = \frac{∆ H_{T_{2}} - 24}{50}$

$\Rightarrow - 0.024 \times 5 = ∆ H_{T_{2}} - 24$

$\Rightarrow ∆ H_{T_{2}} = 22.8 cal / g .$

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