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 $1 mm$ and $T =$ Kelvin temperature. Determine the molar latent heat of vaporization at $80\, K$.

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

Answer 1

Answer: (D)

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

$P=C e^{-\frac{\Delta H_{V}}{R T}}$

$\ln \,P=\ln\, C-\frac{\Delta H_{V}}{R T}$

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

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

$\Delta H_{V}= R [313.7(2.303)+1.40655\, T ]$

at $T=80\, k$

$\Delta H_{V}=(2)[313.7(2.303)+1.40655(80)]$

$\Delta H_{V}=16591$ calorie