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  • (plot of <img alt="\mathrm{\log P}" src="https://entrancecorner.onco

(plot of \mathrm{\log P}, where P= vapour pressure of liquid against \mathrm{\frac{1}{T}} )
The slope of the line will be equal to -1

Option: 1

\mathrm{\Delta H_{\text {vap }}}


Option: 2

\mathrm{-\Delta H_{\text {vap }}}


Option: 3

\mathrm{ -\frac{\Delta H_{\text {Vop }}}{2.303 \mathrm{~K}} }


Option: 4

\mathrm{+\frac{\Delta H_{V a p}}{2.303 R}}


Answers (1)

best_answer

The straight line is in acerdance with clausius clapeyron equation-

\begin{aligned} &\mathrm{ \log \left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{\text {vap }}}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) }\\ &\mathrm{ \log P=\left(-\frac{\Delta H_{\text {vap }}}{2.303 R}\right) \frac{1}{T}+c \text { (constant) (ii) } }\\ &\mathrm{ y=m x+c}\\ \end{aligned}

Hence, slope of the line will be equal to \mathrm{-\frac{\Delta H_{\text {vap }}}{2.303 R}.}

Posted by

Gautam harsolia

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