In the following table is data for the enthalpy of evaporation of nitrogen. Temp

Question

In the following table is data for the enthalpy of evaporation of nitrogen.

Temperature (K) Vapor Pressure (torr)

65   130.5
70 289.5
75 570.8

80 1028

85 1718

The vapor pressure data is plotted above as the natural logarithm of the vapor pressure as a function of 1/temperature. The data was fit to a trend line with a slope of -711.98 K and an intercept of 15.833. What is the value of the value of the heat of vaporization for nitrogen?

*** The answer is ?Hvap = +5.920 kJ/mol but why????? please explain!!

Explanation / Answer

Delta Hvap=-(slope)xR

Delta Hvap=-(-711.98)x(8.314J/mol*K)

this equation only works when its a lnPvap vs. 1/K graph since it uses its slope to determine Delta Hvap

Also answer will be in J/mol so divide by 1000 to get Kj/mol

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