Question

The vapour pressure of a certain liquid is given by the equation: ${\log }_{10}P=3.54595-\frac{313.7}{T}+1.40655{\log }_{10}T$ where $P$ is the vapour pressure $1mm$ and $T=$ Kelvin temperature. Determine the molar latent heat of vaporization at $80K$.

• A. 18591 calories
• B. 17591 calories
• C. 16691 calories
• D. 16591 calories

Answer 1

Answer: (D)

${\log }_{10}P=3.54595-\frac{313.7}{T}+1.40655{\log }_{10}T$

$P=C{e}^{-\frac{\Delta H_V}{RT}}$

$\ln P=\ln C-\frac{\Delta H_V}{RT}$

$\frac{\ln P}{2.303}=3.54595-\frac{313.7}{T}+\frac{1.40655}{2.303}\ln T$

$\frac{\Delta H_V}{R{T}^2}\frac{d(\ln P)}{dt}=+\frac{313.7}{T^2}(2.303)+\frac{1.40655}{T}$

$\Delta H_V=R[313.7(2.303)+1.40655T]$

at $T=80k$

$\Delta H_V=(2)[313.7(2.303)+1.40655(80)]$

$\Delta H_V=16591$ calorie