Screen Reader Note: To calculate percent error, find the absolute value of the difference between experimental value and known value. Take this and divide it the known value. After that, multiply by 100 %. End of note.

Given the known heat of vaporization (delta H vap) of water is 44.01 kJ/mole, find the percent error using the experimentally determined delta H vap of 43.97 kJ/mole. Express with units of % in the units box.

Percent error = (|43.97 kJ/mole - 44.01 kJ/mole| x 100%) / 44.01 kJ/mole = 0.09089 % -> 0.091 %

An unknown liquid has a heat of vaporization of 34.13 kJ/mole. If the normal boiling point is 82, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).

P1 = 760 torr

P2 = ?

T1 = 82 degrees C + 273 = 355 K

T2 = 25 degrees C + 273 = 298 K

delta H = 34.13 kJ/mole x (1000 J / 1 kJ) = 34,130 J/mole

R = 8.314 K/J/mole

ln (P2 / P1) = - (delta H / R) x (1/T2 - 1/T1)

P2 = P1 e^{[-(delta H / R) x (1/T2 - 1/T1)]}

P2 = 760 torr e^{[-( 34,130 J/mole / 8.314 K/J/mole) x (1/298 K -
1/355 K)]}

P2 = 83 torr

Ex.

In a calculator, input:

(760)e^((-34130/8.314)(1/298-1/355))

An unknown liquid has a heat of vaporization of 7.50 kJ/mole. If the
vapor pressure of this liquid at -170 degrees C is 83 torr, what is
the normal boiling point of this liquid **in degrees C**?
HINT: Normal boiling point occurs when the vapor pressure of the
liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).

P1 = 760 torr

P2 = 83 torr

T1 = ?

T2 = -170 degrees C + 273 = 103 K

delta H = 7.50 kJ/mole x (1000 J / 1 kJ) = 7,500 J/mole

R = 8.314 K/J/mole

ln (P2 / P1) = - (delta H / R) x (1/T2 - 1/T1)

ln (83 torr / 760 torr) = - (7,500 J/mole / 8.314 K/J/mole) x (1/103 K - 1/T1)

T1 = 137.85658472 K

T1 = 137.85658472 K - 273.15 K = -135.29

T1 = -135 degrees C

Ex.

- Maybe can use (?): T2 = [ 1/T1 - (( R x ln(P2/P1) / delta H)) ]-1

Screen Reader Note: To calculate percent error, find the absolute value of the difference between experimental value and known value. Take this and divide it the known value. After that, multiply by 100 %. End of note.

Given the known heat of vaporization (delta H vap) of water is 44.01 kJ/mole, find the percent error using the experimentally determined delta H vap of 47.04 kJ/mole. Express with units of % in the units box.

Percent error = (|47.04 kJ/mole - 44.01 kJ/mole| x 100%) / 44.01 kJ/mole = 6.8847989 % -> 6.9 %

An unknown liquid has a heat of vaporization of 39.15 kJ/mole. If the normal boiling point is 85, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).

P1 = 760 torr

P2 = ?

T1 = 85 degrees C + 273 = 358 K

T2 = 25 degrees C + 273 = 298 K

delta H = 39.15 kJ/mole x (1000 J / 1 kJ) = 39,150 J/mole

R = 8.314 K/J/mole

ln (P2 / P1) = - (delta H / R) x (1/T2 - 1/T1)

P2 = P1 e^{[-(delta H / R) x (1/T2 - 1/T1)]}

P2 = 760 torr e^{[-( 39,150 J/mole / 8.314 K/J/mole) x (1/298 K -
1/358 K)]}

P2 = 53.784 torr

P2 = 54 torr

Ex.

In a calculator, input:

(760)e^((-39,150/8.314)(1/298-1/358))

An unknown liquid has a heat of vaporization of 5.50 kJ/mole. If the
vapor pressure of this liquid at -170 degrees C is 111 torr, what is
the normal boiling point of this liquid **in degrees C**?
HINT: Normal boiling point occurs when the vapor pressure of the
liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).

P1 = 760 torr

P2 = 111 torr

T1 = ?

T2 = -170 degrees C + 273 = 103 K

delta H = 5.50 kJ/mole x (1000 J / 1 kJ) = 5,500 J/mole

R = 8.314 K/J/mole

ln (P2 / P1) = - (delta H / R) x (1/T2 - 1/T1)

ln (111 torr / 760 torr) = - (5,500 J/mole / 8.314 K/J/mole) x (1/103 K - 1/T1)

T1 = 147.04434362 K

T1 = 147.04434362 K - 273.15 K = -126.10565638

T1 = -126 degrees C^{}