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
I don't think you have enough information.
dG = dH - TdS
At boiling, dG = 0
Then TdS = dH and
dS = dH/T = 540/373 but that gives you cal/g and you want cal/mol. Without knowing the molar mass (I assume this is water but the problem doesn't say that), I can't convert from cal/gram to cal/mol.
26calk/mol
To calculate the entropy increase in the evaporation of a liquid, we can use the equation:
ΔS = q/T
Where ΔS is the entropy change, q is the heat absorbed or released, and T is the temperature in Kelvin.
In this case, we have the heat of vaporization given in calories per gram. However, we need to convert this to calories per mole.
Since we are given the heat of vaporization at 100 degrees Celsius, which is the boiling point, we need to convert this temperature to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 100 + 273.15 = 373.15 K
Now, we need to convert the heat of vaporization from calories per gram to calories per mole. We know that the molar mass of the substance is required for this conversion.
Let's assume the molar mass (M) of the substance is X g/mol.
Using the heat of vaporization (H) given, which is 540 cal/g, the heat of vaporization per mole (H') can be calculated as:
H' = H * M
Now, we can calculate the entropy change (ΔS):
ΔS = H' / T
Substituting the values:
ΔS = (540 cal/g * X g/mol) / 373.15 K
Simplifying the equation, the units of grams will cancel out and we will be left with:
ΔS = (540 * X) / 373.15 cal/K
Note: If you know the molar mass of the substance, you can substitute it as X in the equation above to find the entropy increase.