Vapor pressure of methanol at 25 ° C has to be calculated. Concept Introduction: Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid by “ Clausius – Clapeyron ” equation as follows – ln P = -ΔH vap R ( 1 T ) + C Where T = Temperature P = vapor pressure of the liquid at temperature T ΔH vap = Molar heat of vaporization R = Universal Gas constant . At two different temperature and pressure, the equation is rewritten as – ln P 1 P 2 = -ΔH vap R ( 1 T 2 - 1 T 1 ) + C Where T 1 and T 2 are two different Temperature P 1 and P 2 are two different Pressure ΔH vap = Molar heat of vaporization R = Universal Gas constant .
Vapor pressure of methanol at 25 ° C has to be calculated. Concept Introduction: Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid by “ Clausius – Clapeyron ” equation as follows – ln P = -ΔH vap R ( 1 T ) + C Where T = Temperature P = vapor pressure of the liquid at temperature T ΔH vap = Molar heat of vaporization R = Universal Gas constant . At two different temperature and pressure, the equation is rewritten as – ln P 1 P 2 = -ΔH vap R ( 1 T 2 - 1 T 1 ) + C Where T 1 and T 2 are two different Temperature P 1 and P 2 are two different Pressure ΔH vap = Molar heat of vaporization R = Universal Gas constant .
Solution Summary: The author explains the Vapor Pressure of methanol at 25°C has to be calculated.
Vapor pressure of methanol at
25°C has to be calculated.
Concept Introduction:
Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid by “Clausius – Clapeyron” equation as follows –
At two different temperature and pressure, the equation is rewritten as –
lnP1P2=-ΔHvapR(1T2-1T1)+CWhereT1 and T2 are two different Temperature P1 and P2 are two different PressureΔHvap=MolarheatofvaporizationR=UniversalGas constant.
Draw product A, indicating what type of reaction occurs.
NH2
F3C
CF3
NH
OMe
NH2-NH2, ACOH
A
Photochemical smog is formed in part by the action of light on nitrogen dioxide. The wavelength of radiation absorbed by NO2 in this reaction is 197 nm.(a) Draw the Lewis structure of NO2 and sketch its π molecular orbitals.(b) When 1.56 mJ of energy is absorbed by 3.0 L of air at 20 °C and 0.91 atm, all the NO2 molecules in this sample dissociate by the reaction shown. Assume that each absorbed photon leads to the dissociation (into NO and O) of one NO2 molecule. What is the proportion, in parts per million, of NO2 molecules in this sample? Assume that the sample behaves ideally.
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