The variation of vapor pressure p (in units of mm Hg) of benzene with temperatur

Question

The variation of vapor pressure p (in units of mm Hg) of benzene with temperature in the range of 0 T 42 °C can be modeled with the equation:

ln(p)= (K /0.0101 a)-b

where a=34172 and b=7.9622 are material constants and T is absolute temperature (K).
Write a program that calculates the pressure at various temperatures and displays the results in a three column table. The first column must be temperatures in °C, the second column, temperatures in K and the third column is the corresponding pressures in mm Hg. Include title and header information for each column.
Note: (i) Use an increment of 2 degrees in your temperature vector.

(ii) You may have to use format short g or format short e to format pressures.

Explanation / Answer

ln(p)= (K /0.0101 a)-b

temperatures let us take 0o, 10o, 20o

corresponding temperature in kelvin

T = 273 + t

0o = 273 K

10o = 273+10 = 283 K

20o = 273 +20 = 293K

equation :

ln(p)= (K /0.0101 a)-b

at 0o = 273 K

lnP = ( 273 / 0.0101 x 34172 ) - 7.9622

lnP       = -7.17

P = e^-7.17

P = 7.69 x 10^-4 mmHg

(ii)

at 10o = 283 K

lnP = ( 283 / 0.0101 x 34172 ) - 7.9622

lnP       = -7.14

P = e^-7.14

P = 7.92 x 10^-4 mmHg

(iii)

at 20o = 293 K

lnP = ( 293 / 0.0101 x 34172 ) - 7.9622

lnP       = -7.11

P = e^-7.11

P = 8.17 x 10^-4 mmHg

temperature (oC) temperature (oC) pressure (mmHg) 0 273 7.69 x 10^-4 10 283 7.92 x 10^-4 20 293 8.17 x 10^-4

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