calculate the heat f iodine vapours & solid are 0.031 & 0.055 cal/g respectively. If heat of sublimation f iodine is 24 cal/ g at 200°C .wat is its value at 260°C??

To calculate the heat required to convert iodine from solid to vapor at a given temperature, you can use the following equation:

ΔH(sublimation) = ΔH(vaporization) + ΔH(fusion)

where ΔH(sublimation) is the heat of sublimation, ΔH(vaporization) is the heat of vaporization, and ΔH(fusion) is the heat of fusion.

Given that the heat of sublimation of iodine is 24 cal/g at 200°C, we need to find the heat at 260°C.

To begin, we need to calculate the heat of vaporization at 200°C. We can use the heat of fusion and the heat of vaporization to find it:

ΔH(vaporization) = ΔH(sublimation) - ΔH(fusion)

ΔH(vaporization) = 24 cal/g - 0.055 cal/g

ΔH(vaporization) = 23.945 cal/g (approximately)

Now, we can find the heat of vaporization at 260°C by using the specific heat capacity of iodine vapor:

ΔH(vaporization at 260°C) = ΔH(vaporization at 200°C) + heat of iodine vapor × (temperature change)

Given that the heat of iodine vapor is 0.031 cal/g, and the temperature change is 260°C - 200°C = 60°C:

ΔH(vaporization at 260°C) = 23.945 cal/g + 0.031 cal/g × 60°C

ΔH(vaporization at 260°C) = 23.945 cal/g + 1.86 cal/g

ΔH(vaporization at 260°C) = 25.805 cal/g (approximately)

Therefore, the heat required to convert iodine from solid to vapor at 260°C is approximately 25.805 cal/g.