Question

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.

Answer 1

Answer: 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:

$\Delta S_{vap}=n\times \frac{\Delta H_{vap}}{T}$

where,

$\Delta S_{vap}$ = Entropy change of vaporization = ?

n = number of moles of pentane = 0.34 moles

$\Delta H_{vap}$ = molar heat of vaporization = 25.8 kJ/mol

T = temperature of the system = $36.1^oC=[36.1+273]K=309.1K$

Putting values in above equation, we get:

$\Delta S_{vap}=\frac{0.34mol\times 25.8kJ/mol}{309.1K}\\\\\Delta S_{vap}=0.0284kJ/mol$

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