Propane (C3H8) has a normal boiling point of -42.0°C and a heat of Vaporization (Hvapo) of 19.04Kj/mol. What is the Vapor Pressure of propane at 25.0°C?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

According to Clausius Clapeyron equation:

ln(P2/P1) = (-∆Hvap)/R × (1/T2 - 1/T1),

where T1 - boiling point (-42°C = 315.15 K), T2 - temperature (25°C = 298.15 K), ∆Hvap - heat of vaporizayion (19.04 Kj/mol), R - gas constant (8.314 J/mol K), P1 - atmospheric pressure (1 atm = 101.325 kPa), P2 - vapor pressure (?).

From here:

ln(P2) - ln(P1) = (-∆Hvap)/R × (1/T2 - 1/T1)

ln(P2) = (-∆Hvap)/R × (1/T2 - 1/T1) + ln(P1)

P2 = e[(-∆Hvap)/R × (1/T2 - 1/T1) + ln(P1)]

P2 = 939 kPa

Answer: 939 kPa

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 ∘C. At what temperature does benzene boil when the external pressure is 410 torr?

The external pressure of the benzene is: Pb = 410 torr = 54662.20 PbThe molar enthalpy of vaporization of benzene is: ΔHa =30.72 Kj/mol = 30720 J/molThe normal boiling...

1 Answers 1 views
Propane, C3H8, used in gas barbeques, burns in oxygen to produce carbon dioxide and water according to the following equation: C3H8 + 5 O2

C3H8 +5O2 = 3CO2 +4H2O.

1 Answers 1 views
A certain liquid has a normal boiling point of 338.15k. using Trouton's rule 1. estimate the vapor pressure at 325.15k 2.estimate the boiling point at a pressure of 0.267bar

Lvap / T BOILLING = 85 Lvap = ( 85 × 338.15 ) j = 28370 j = 28.370 kj.

1 Answers 1 views
The vapor pressure of 1-propanol is 10.0 torr at 14.7 °C. Calculate the vapor pressure at 52.8 °C.Given: Heat of vaporization of 1-propanol = 47.2 kJ/mol

Vapor pressure=34.2atm

1 Answers 1 views
A fuel tank used in a barbeque contains 1.09 x 10-1 moles of propane, C3H8(g). How many molecules of propane are in the tank?

Solution:Moles of propane = 1.09×10-1 moles = 0.109 molesOne mole of any substance contains 6.022×1023 atoms/molecules.Hence,Number of propane molecules = 0.109 mol × (6.022×1023 molecules / 1 mol) = 6.56×1022 moleculesAnswer: 6.56×1022 molecules...

1 Answers 1 views
Calculate the heat required to vaporize 7.87 g of benzene at its normal boiling point. Heat of vaporization (benzene) = 30.7 kJ/mol

Find the amount of benzene (Mbenzene = 78 g/mol) ν = 7.87/78 = 0.1 (mole) Find the heat required: Q = 0.1 × 30.7 = 3.07 (kJ)

1 Answers 1 views
Calculate the heat required to vaporize 7.87 g of benzene at its normal boiling point. Heat of vaporization (benzene) = 30.7 kJ/mol Heat = kJ

Solution: Molecular weight of benzene: 78.11 g/mol Amount of benzene: 7.87 / 78.11 = 0.1 mol Energy needed (heat): 0.1 ∙ 30.7 = 3.07 kJ Answer: Heat = 3.07 kJ

1 Answers 1 views
The standard (1 bar) boiling point of acetone is 56.1 C, and its enthalpy of vaporization is 29.1 kJ/mol. Calculate the vapor pressure of acetone at 43.5 C.

ln(p2 / p1)=- (∆Hvap/R) x (1/T2 - 1/T1)ln(p2/1)= -(29.1/8.314) x (1/329.25 - 1/316.65)p2=1.000423 bar

1 Answers 1 views
Propane (C3H8) has a normal boiling point of -42.0°C and a heat of Vaporization of 19.04Kj/mol. What is the Vapour Pressure of propane at 25.0°C?

T1 = -42.0 + 273.15 = 231.15 KT2 = 25 + 273.15 = 298.15∆Hvap = 19.04 kJ/mol = 19040 J/molAt normal boiling point, the vapor pressure is equal to 1...

1 Answers 1 views
A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure is 1.50 atm. Calculate the partial pressures of the gases.

p(CH4) = 0.55 atmp(C2H6) = 0.44 atmp(C3H8)= 0.51 atm

1 Answers 1 views
X