Find the increase in entropy of 1.00 kg of liquid nitrogen that starts at its boiling temperature, boils, and warms to 20.0 º C at constant pressure.

We calculate the change in entropy for both when nitrogen boils and changes to the vapor phase and when this nitrogen in gaseous form is further heated to 20^oC. We add both these changes in entropies to get the total change in entropy.

Latent heat of vaporization of nitrogen L=199kJ/kg

Final temperature of nitrogen T_h=20 ̊ C = 293 K

Boiling point of nitrogen T_c= 77.36 K

Specific heat of nitrogen c=1.04 kJ/kg ̊ C

Change in entropy

\[\Delta S= \frac{Q}{{\bf{T}}}\]

Here,

\[\Delta S\] - change in entropy.

Q - heat transferred.

T- temperature.

Energy is transferred to nitrogen when it is boiling

Q =mL

= 1 kg×199000 J/kg

= 199000 J

Change in entropy of nitrogen

\begin{aligned}\Delta{S_1}=\frac{Q}{{{T_c}}}\\=\frac{{199000\;{\rm{J}}}}{{77.36\;{\rm{K}}}}\\= 2572.39\;{\rm{J/K}} \end{aligned}

Temperature changes when nitrogen is heated to 20^oC

∆T = 20-(-195.8)

= 215.8 ̊ C

Energy is transferred to nitrogen when it is heating

Q = mc ∆T

= 1kg ×1040 J/kg. ̊ C ×215.8 ̊ C

= 224000 J

Average temperature for nitrogen heating

\begin{aligned}{T_{avg}} = \frac{{({T_c} + {T_h})}}{2}\\= \frac{{(77.36\;{\rm{K}} + 293\;{\rm{K}})}}{2}\\= 185.18\;{\rm{K}}\end{aligned}

Change in entropy of gaseous nitrogen

\begin{aligned}\Delta{S_2}=\frac{Q}{{{T_{avg}}}}\\=\frac{{224000\;{\rm{J}}}}{{185.18\;{\rm{K}}}}\\= 1209.63 {\rm{J/K}}\end{aligned}

Total change in entropy

\begin{aligned}\\Delta {S_{total}}= \Delta {S_1} + \Delta {S_2}\\= (2572.39 + 1209.63)\;{\rm{J/K}}\\= 3782.02 {\rm{J/K}}\end{aligned}

Therefore, the total increase in entropy for nitrogen is 3782.02 J/K.