## Mention brand new Pattern

Erratic substances possess reduced boiling hot factors and apparently poor intermolecular relations; nonvolatile ingredients have large boiling hot facts and you can seemingly solid intermolecular relationships.

The new great upsurge in steam tension having broadening temperatures in Shape “The fresh new Steam Challenges of a lot Liquid due to the fact a purpose of Heat” lets us explore sheer logarithms to share with you the nonlinear relationship just like the good linear you to definitely. 9 “Important Event six”.

ln P = ? ? H vap Roentgen ( step one T ) + C Formula having a straight line : y = yards x + b

where ln P is the natural logarithm of the vapor pressure, ?H_{vap} is the enthalpy of vaporization, R is the universal gas constant [8.314 J/(mol·K)], T is the temperature in kelvins, and C is the y-intercept, which is a constant for any given line. A plot of ln P versus the inverse of the absolute temperature (1/T) is a straight line with a slope of ??H_{vap}/R. Equation 11.1, called the Clausius–Clapeyron equation A linear relationship that expresses the nonlinear relationship between the vapor pressure of a liquid and temperature: ln P = ? ? H vap / R T + C , where P is pressure, ? H vap is the heat of vaporization, R is the universal gas constant, T is the absolute temperature, and C is a constant. The Clausius–Clapeyron equation can be used to calculate the heat of vaporization of a liquid from its measured vapor pressure at two or more temperatures. , can be used to calculate the ?H_{vap} of a liquid from its measured vapor pressure at two or more temperatures habbo arama. The simplest way to determine ?H_{vap} is to measure the vapor pressure of a liquid at two temperatures and insert the values of P and T for these points into Equation 11.2, which is derived from the Clausius–Clapeyron equation:

ln ( P dos P step one ) = ? ? H v a good p Roentgen ( 1 T dos ? step one T 1 )

Conversely, if we know ?H_{vap} and the vapor pressure P_{1} at any temperature T_{1}, we can use Equation 11.2 to calculate the vapor pressure P_{2} at any other temperature T_{2}, as shown in Analogy 6.

## Example 6

From these data, calculate the enthalpy of vaporization (?H_{vap}) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)

A Use Equation 11.2 to obtain ?H_{vap} directly from two pairs of values in the table, making sure to convert all values to the appropriate units.

A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation 11.1 and find the value of ?H_{vap} from the slope of the line, an alternative approach is to use Equation 11.2 to obtain ?H_{vap} directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvins because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T_{1}) and 120.0°C (T_{2}) into Equation 11.2 gives

ln ( 0.7457 torr 0.0888 torr ) = ? ? H vap 8.314 J/(mol · K) [ 1 ( 120 + 273 ) K ? 1 ( 80.0 + 273 ) K ] ln ( 8.398 ) = ? ? H vap 8.314 J · mol ? step 1 · K ? 1 ( ? dos.88 ? ten ? cuatro K ? step 1 ) 2.13 = ? ? H vap ( ? 0.346 ? ten ? cuatro ) J ? 1 · mol ? H vap = 61,eight hundred J/mol = 61 .4 kJ/mol

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