Please answer my chemistry question below which is abt finding heat of vaporization & normal boiling point?

Question

The vapor pressure of nitrogen at various temperatures is provided in the table below. Using this data, calculate:
A) The heat of vaporization of nitrogen.
B) The normal boiling point of nitrogen.  
Temperature (K) | Pressure (torr)
-----------------|----------------
65              | 130.5
70              | 289.5
75              | 570.8
80              | 1028
85              | 1718  
Please provide a complete solution with detailed steps, not just the final answers.

Anonymous · Accepted Answer

The vapor pressure of nitrogen at several different temperatures is shown below. Use the data to find A) heat of vaporization & B) normal boiling point. 
When the vapor pressure = atmospheric pressure, the liquid boils!
Standard atmospheric pressure = 760 torr
Temperature (K), Pressure (torr)
65 , 130.5
70 , 289.5
75 , 570.8
80 , 1028
85 , 1718
You need an equation for the graph of the data above, see web site below! 
https://eee.uci.edu/programs/gchem/05MANPTgasVPliq...
ln P = (-1  Heat of vaporization ÷ R) (1/T) + C
R = 8.3145 J/(mole * °K)
T = temperature in °K
ln P = (-1  Heat of vaporization ÷ 8.3145) (1/T) + C
Use several values of T and P to determine the value of Heat of vaporization
ln 130.5 = (-1  Heat of vaporization ÷ 8.3145) (1/65) + C
C = ln 130.5 – (-1  Heat of vaporization ÷ 8.3145) (1/65)
ln 570.8 = (-1  Hv ÷ 8.3145) (1/75) + ln 130.5 – (-1  Hv ÷ 8.3145) (1/65)
Subtract ln 130.5 from both sides
ln 570.8 – ln 130.5 = (-1  Hv ÷ 8.3145) (1/75) – (-1  Hv ÷ 8.3145) (1/65)
ln 570.8 – ln 130.5 =  Hv  (-1 ÷ 8.3145) (1/75) – (-1 ÷ 8.3145)* (1/65)
ln 570.8 – ln 130.5 = ( Hv /8.3145) * [(-1/75) – (-1/65)]
(6.347 – 4.871)  8.3145 = Hv  [(-1/75) – (-1/65)]
Hv = (1.476 * 8.3145) ÷ 0.00205
Hv = 5986.44
C = ln 130.5 – (-1  5986.44 ÷ 8.3145) (1/65)
C = ln 130.5 – (-11.077)
C = 15.95
Then use the equation to determine the Temperature when P = 760 torr  
ln P = (-720) * (1/T) + 15.95
P = 760
ln 760  = (-720) * (1/T) + 15.95 
ln 760 – 15.95 = (-720) * (1/T)
(ln 760 – 15.95) ÷ (-720) = (1/T)
T = 77.3 °K
http://en.wikipedia.org/wiki/Nitrogen
BP = 77.36 °K
WOW

Anonymous · Answer

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RE:
Please answer my chemistry question below which is abt finding heat of vaporization & normal boiling point?
The vapor pressure of nitrogen at several different temperatures is shown below. Use the data to find A) heat of vaporization & B) normal boiling point. 
Temperature (K), Pressure (torr)
65 , 130.5
70 , 289.5
75 , 570.8
80 , 1028
85 , 1718
*PLEASE GIVE ME THE COMPLETE SOLUTION & NOT JUST...

Kameron Wyman

Please answer my chemistry question below which is abt finding heat of vaporization & normal boiling point?

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