The molar heat of vaporization of pentane is 25.8 kJ·mol−1, and the boiling point of pentane is 36.1°C. Calculate the value of ΔvapS for the vaporization of 0.34 mole of pentane.

The entropy change of vaporization of pentane is 0.0284 kJ/mol

Explanation:

To calculate the entropy change for vaporization of pentane, we use the equation:

where,

= Entropy change of vaporization = ?

n = number of moles of pentane = 0.34 moles

= molar heat of vaporization = 25.8 kJ/mol

T = temperature of the system =

Putting values in above equation, we get:

Hence, the entropy change of vaporization of pentane is 0.0284 kJ/mol