Question

What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C∘C and ΔHvapΔHvap = 28.0 kJ/molkJ/mol .

Answer 1

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$= initial pressure at $26.5^oC$ = ?

$P_2$ = final pressure at $-5.1^oC$ = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$= initial temperature = $26.5^oC=273+26.5=299.5K$

$T_2$ = final temperature =$-5.1^oC=273+(-5.1)=267.9K$

Now put all the given values in this formula, we get

$\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]$

$\log (\frac{P_1}{100})=-0.576$

$\frac{P_1}{100}=0.265$

$P_1=26.5mmHg$

Thus the vapor pressure of $CS_2CS_2$ in mmHg at 26.5 ∘C is 26.5