It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (MW: 46.07 g/mol). What will be ΔSsys for the vaporization of 8.00 g of ethanol at 79.6 °C

Explanation:

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In this case, given the required energy to vaporize 1.00 mol of ethanol as the enthalpy of vaporization:

We can compute the entropy of the system for the vaporization of 8.00 g of ethanol, by first computing the moles:

And then setting up the following expression:

Whereas the temperature is in kelvins; thus, we obtain:

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2b2t

explanation:

2b2t