Question

Ethanol's enthalpy of vaporization is 38.7kJmol^-1at its normal boiling

point, 78⁰. Calculate$\mathit{q}\mathbf{,}\mathit{w}\mathbf{,}\mathit{\Delta }\mathit{U}\mathbf{,}\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{y}\mathbf{s}}$, and $\mathbf{△}\mathit{G}$when 1.00molethanol is vaporized reversibly at 78⁰and 1 atm. Assume that the vapor is an ideal gas and neglect the volume of liquid ethanol relative to that of its vapor.

Short answer

(a)The amount of heat,q=+38.7kJ

(b)The work done,w=-2.92kJ

(c)The internal energy,$△U=+35.8kJ$

(d)The change of entropy$△S_{sys}=110J{K}^{-1}$

(e)The Gibbs free energy, $△G=0$

Step-by-step solution

Given data

The heat of vaporization,$△H_{vap}=38.7kJmo{l}^{-1}$

The temperature of boiling point,data-custom-editor="chemistry" ${\mathrm{T}}_{\mathrm{boiling}}=78°\mathrm{C}$

Concept Gibbs free energy

At constant temperature and pressure, Gibbs free energy is a thermodynamic potential that can be used to compute the maximum work that a thermodynamically closed system can perform.

Calculation of volume

Calculation of work done

The work done is calculated with the help of the formula:

$w=-P\Delta V$

It is known that,$1\mathrm{Latm}=101.325\mathrm{J}$

where w is work done, $△\mathrm{V}$is a change in volume.

Put the value of the given data in the above equation.

$\begin{array}{c}w=1atm\times 28.8L\\ =-28.8Latm\times \frac{101.325J}{1Latm}\\ =-2.92\times {10}^{3}S\times \frac{1kJ}{{10}^{3}J}\\ =-2.92kJ\end{array}$

Calculation of internal energy 

The internal energy is calculated with the help of the formula:

$△U=q+w$

where w is work done, $△U$is the internal energy.

Put the value of the given data in the above equation.

$\begin{array}{c}\Delta U=38.7kJ+(-2.92kJ)\\ =+35.8kJ\end{array}$

Calculation of change of entropy for the system

Calculation of Gibbs free energy