What is the vapor pressure of propane at 10.0 ∘C?
At -42.0 C (231 K) the vapor pressure is 1 atm. So, using the Clausius-Clapeyron Equation: ln (P2 / P1) = -(ΔHvap/R) (1/T2-1/T1) ln (P2/1) = -(19040 J/mol / 8.314 J/molK) (1/283K - 1/231K) ln(P2) = 1.822 P2 = 6.18 atm
Propane has a normal boiling point of -42.0 C and a heat of vaporization of 19.04KJ/mol?
You have to use Clausius-Clayperon eq: ln pv = [ - D(H)vap / (R x T) ] + C where R = 8.31 J/oK.mol and the T in oK pv = vapor pressure For the 2 different conditions: p1 = 760 mm Hg T1 = 231oK p2 = ? T2 = 298oK Then: ln(p2/p1) = - D(H) / R [ (1 / T2) – (1 / T1) ] ln(p2/760) = [ - 19,040.00 / 8.31 ] x [ ( 1 / 298) – ( 1 / 231) ] ln(p2/760) = - 2,291 [ 3.36 x (– 4.33) ] x 10^(-3) [ ln(p2/760) ] = 2,291 x 0.97 x 10^(-3) = 2.22 [ p2 / 760 ] = 9.21 p2 ~ 7,000 mm Hg ~ 9,21 atm ~ 135.4 psia
Propane has a normal boiling point of -42.0 ∘C and a heat of vap (ΔHvap) of 19.04 kJ/mol. What is the vapor pressure of propane at 35.0∘C?
ln (P2/P1) = (ΔHvap/R)(1/T1−1/T2) ln (P2/1) = (19.04 / 8.314) (1/231 - 1/309) ln (P2) = 0.00741 P2 = e^0.00741 = 1.074 atm
Propane has a normal boiling point of -42.0 ∘C and a heat of vaporization (ΔHvap) of 19.04 kJ/mol. What is the?
Can someone explain to me the proper way to go about this? I'm not quite getting it. Propane has a normal boiling point of -42.0 ∘C and a heat of vaporization (ΔHvap) of 19.04 kJ/mol. What is the vapor pressure of propane at 10.0∘C?
Clausius Clapeyron / increasing vapor pressure helppp plz?
1) Solid iodine has a vapor pressure of 1.0 mmHg at 39°C. How many moles of iodine will sublime into a 500. mL flask at this temperature? If the volume of the flask is doubled at constant temperature, what will happen to the equilibrium vapor pressure of I2? (Assume some solid I2 is always present in the container.) A.2.1 x 10-4 mol; vapor pressure increases B.2.0 x 10-2 mol; vapor pressure increases C.2.6 x 10-5 mol; no change in vapor pressure D.2.1 x 10-4 mol; no change in vapor pressure E.2.6 x 10-5 mol; vapor pressure decreases 2) Find the temperature at which ethanol boils on a day in the mountains when the barometric pressure is 547 mmHg. (Given: The heat of vaporization of ethanol is 39.3 kJ/mol; the normal boiling point of ethanol is 78.3°C.) A.76.5°C B.69.9°C C.10.0°C D.77.9°C E.74.6°C 3)The vapor pressure of ethanol is 400 mmHg at 63.5°C. Its molar heat of vaporization is 39.3 kJ/mol. What is vapor pressure of ethanol, in mmHg, at 34.9°C? A.1,510 mmHg B.100 mmHg C.200 mmHg D.0.0099 mmHg E.4.61 mmHg 4)Each of the following substances is a liquid at -50°C. Place these liquids in order of increasing vapor pressure. dimethyl ether (CH3OCH3), propane (C3H8), ethanol (CH3CH2OH) A.ethanol < propane < dimethyl ether B.ethanol < dimethyl ether < propane C.propane < dimethyl ether < ethanol D.dimethyl ether < ethanol < propane E.propane < ethanol < dimethyl ether
The normal boiling point of decane is 174°C and the ΔHvap is 39.58 kJ/mol. What would the vapor pressure of decane be at 25°C?
from the clausius clapeyron equation .. ln(P1 / P2) = (dHvap / R) x (1/T2 - 1/T1) the idea being .. if you have 2 sets of pressure / temperature points .. they are related by dHvap / R in this problem .. P1 = 1 atm .. T1 = 174°C = 447.15K .. P2 = ? .. T2 = 298.15K .. dHvap = 39580 J/mol .. R = 8.314 J/molK rearranging that above equation .. P1 / P2 = exp[(dHvap / R) x (1/T2 - 1/T1)] .. P2 = P1 / exp[(dHvap / R) x (1/T2 - 1/T1)] solving .. P2 = 1atm / exp[ ((39580 J/molK) / (8.314 J/mol)) x (1/298.15K - 1/447.15K) ] calculating .. P2 = 0.00489 atm (3.72mmHg)
Can Turboflame Phoenix lighters be refilled with propane gas?
Propane has a much lower boiling point than butane, that means it is a much higher pressure gas at normal temperatures. You'd have to change the jet, which you can't, even if the body was safe at the higher pressure. At room temperature the vapor pressure of butane is around 30psi and propane is over 90psi. Don't do it.
How is propane both a liquid and a gas?
See if you can find a refillable propane cigarette lighter or lighter for the logs in your fireplace. You will notice that it has a transparent tube. if you hold the tube vertically, you can see liquid in the bottom and vapor in the top. Both are propane. At room temperature and pressure propane is a vapor. At elevated pressure, propane is a liquid. As you use the lighter the propane vapor escapes , is lit, and burned away. Also, the pressure is reduced and the liquid propane evaporates into more vapor. You will see the top of the vertical tube to show more and more vapor and the bottom of the tube will show less and less liquid propane. This is not so much related to the lesser pressure inside the lighter. This is more a matter of the propane has been consumed and there's less liquid in the lighter.Technically, a gas is a chemical above its critical temperature. That means no amount of compression or pressure can convert a gas to a liquid. Oxygen is a good example of a gas at room temperature. No amount of pressure can convert it to a liquid. But when you go to a hospital, in the back, you will find a gigantic cylinder. it's filled with liquid oxygen and it's refrigerated so the temperature is extremely low. It gets the temperature below the critical temperature of oxygen. Once the temperature is so low, below the critical temperature, the oxygen can be compressed into a liquid.