Respuesta :
Answer:
Answers:
Option B
Option A.
Explanation:
For part A
The applicable equation is the Clausius-Clapeyron equation. The equation relates the vapor pressure and temperature like this:
[tex]ln(\frac{P_{2} }{P_{1} } = \frac{\deltaH_{vap} }{R} * (\frac{1}{T_{2} } - \frac{1}{T_{1}) }[/tex]
We know that P₁ = 6.63 torr
R = 8.314 kj/mol
T₁ = 106.5⁰ = 379.65 K
T₂ = 38⁰ = 311.15 K
ΔH = 43 300J/mol
using the equation to solve for P₂, we get 67.8 torr
For part B
Water is a polar molecule. The bonds in the water are hydrogen bonds; water forms strong hydrogen bonds between the molecules. However, the gases in the air are non polar. Nitrogen and argon are inert gases - nonreactive. Nitrogen has a triple bond N≡N and argon is inert. Oxygen is a bi-atomic gas, O₂.
Hence the only forces in the water with the gases are dipole-induced dipole forces.
(A) The pressure of gasoline at 38 degree Celsius has been 67.8 torr. Thus, option B is correct.
(B) The dissolution of gases in water has been induced by the dipole induced dipole interaction. Thus, option A is correct.
The vapor pressure has been the pressure exerted by the gaseous molecules in the mixture. It has been dependent on the temperature of the solution.
Computation for the vapor pressure
(A) The vapor pressure with respect to temperature has been given by Clausius-Clapeyron equation as:
[tex]ln(\dfrac{P_2}{P_1})=\dfrac{\Delta H}{R}\;(\dfrac{1}{T_2} -\dfrac{1}{T_1} )[/tex]
Where, the known vapor pressure of gasoline is [tex]P_1=6.63\;\rm torr[/tex]
The enthalpy of vaporization of gasoline is [tex]\Delta H=43.3\;\rm kJ/mol\\\Delta \textit H=43,300\;J/mol[/tex]
The initial temperature of gas is [tex]T_1=106.5^\circ \text C\\T_1=379.65\;\text K[/tex]
The final temperature of the gas is [tex]T_2=38^\circ \text C\\T_1=311.15\;\text K[/tex]
The value of Boltzmann constant, [tex]R=8.314\;\rm J/mol.K[/tex]
Substituting the values for the final pressure ([tex]P_2[/tex]) of the gas:
[tex]ln(\dfrac{P_2}{6.63})=\dfrac{43,300}{8.314}\;(\dfrac{1}{311.15} -\dfrac{1}{379.65} )\\\\ln(\dfrac{P_2}{6.63})=3.020\\\\P_2=67.8\;\rm torr[/tex]
The pressure of gasoline at 38 degree Celsius has been 67.8 torr. Thus, option B is correct.
(B) The polar molecules have been composed of partial charged. Water is a polar molecule with the partial positive hydrogen and partial negative oxygen.
The dissolution of gases in the polar molecule has been induced by the dipole induced dipole interaction, as gases has been nonpolar gases.
Thus, option A is correct.
Learn more about vapor pressure, here:
https://brainly.com/question/25356241
