Question

The vapour pressure of a certain liquid is given by the equation:
$lo{g}_{10}P=3.54595+\frac{313.7}{T}+1.40655lo{g}_{10}T$, where, P is the vapour pressure in mm and T is temperature in K. The molar latent heat of vaporisation as a function of temperature and its value in cal at 80 K is :

• A. $597$
• B. $778.56$
• C. $1056.24$
• D. $1194$

Answer 1

The correct option is D $1194$

$lo{g}_{10}P=3.54595+\frac{313.7}{T}+1.40655lo{g}_{10}T$
At 80 K
$lo{g}_{10}P=3.54595+\frac{313.7}{80}+1.40655lo{g}_{10}\times 80=10.144$
At 100 K
$lo{g}_{10}{P}^{\prime }=3.54595+\frac{313.7}{100}+1.40655lo{g}_{10}\times 100=9.49605$
$log\frac{P^{\prime }}{P}=\frac{\Delta H}{2.303R}(\frac{1}{T}-\frac{1}{T^{\prime }})$
$9.49605-10.144=\frac{\Delta H}{2.303\times 2}(\frac{1}{80}-\frac{1}{100})$
$\Delta H=-1194cal$
Hence, the molar latent heat of vaporization is 1194 cal.