## Related questions with answers

Carbon disulfide, $\mathrm{CS}_2$, is a volatile, flammable liquid. It has a vapor pressure of $400.0 \mathrm{mmHg}$ at $28.0^{\circ} \mathrm{C}$ and $760.0 \mathrm{mmHg}$ at $46.5^{\circ} \mathrm{C}$. What is the heat of vaporization of this substance?

Solution

VerifiedThe vapor pressure of a substance depends on temperature. We can calculate values of the vapor pressure at different temperatures by using the Clausius-Clapeyron equation which relates the vapor pressure, the temperature, and the heat of vaporization of a substance.

$\text{ln}\dfrac{P_{2}}{P_{1}}=\dfrac{\Delta{H}_{vap}}{R}(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$

Where $\textit{P}_{1}$ and $\textit{P}_{2}$ are the values of the vapor pressure at **absolute** temperatures $\textit{T}_{1}$ and $\textit{T}_{2}$.

$\Delta{H}_{vap}$ is the heat of vaporization, and $\textit{R}$ is the molar gas constant which equals 8.314 J/ K$\cdot$mol.

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