Calculate the entropy increase in the evaporation of 1 mole of a liquid when it boils at 100 degree centigrade having heat of vaporisation at 100 degree as 540cals\gm
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To calculate the entropy increase during the evaporation of 1 mole of a liquid, we need to use the equation:
ΔS = q/T,
where ΔS is the entropy change, q is the heat transferred, and T is the temperature in Kelvin.
First, let's convert the heat of vaporization from calories to joules. Since 1 cal = 4.184 J:
Heat of vaporization (q) = 540 cal/g * 4.184 J/cal = 2254.16 J/g.
Since we have 1 mole of the liquid, we need to find the molar mass to convert the heat of vaporization from grams to moles. Let's assume the molar mass of the liquid is M g/mol.
So, the heat of vaporization in joules per mole is:
q_per_mole = 2254.16 J/g * (1 g/M) = 2254.16 J/mol.
Next, we need to calculate the temperature in Kelvin. Since the boiling point is given as 100 degrees Celsius, we need to add 273.15 to convert it to Kelvin.
Therefore, T = 100 + 273.15 = 373.15 K.
Now, we can substitute the known values into the entropy change equation:
ΔS = q_per_mole / T = 2254.16 J/mol / 373.15 K.
Calculating this, the entropy increase during the evaporation of 1 mole of the liquid is approximately equal to the value obtained from the above equation.