Respuesta :
Answer:
238.3 torr
Explanation:
We'll use the Clausius-Clapeyron Equation which allows us to calculate the vapour pressure of a liquid at a certain temperature, given that the enthalpy of vaporization and the vapor pressure at another temperature is known.
㏑[tex](\frac{P_{1} }{P_{2} } )[/tex] = [tex]\frac{H_{vap} }{R}[/tex] [tex](\frac{1}{T_{1} } - \frac{1}{T_{2} } )[/tex]
T₁ (temperature)= (50.0+ 273.15) K = 323.15 K (convert temperature to kelvin)
P₁ (vapor pressure) = ???
T₂ (temperature) = (78.4+ 273.15) K = 351.55 K (convert temperature to kelvin)
P₂ (vapor pressure) = 760 torr
R = Ideal gas constant = 8.314 J⋅K ⁻¹mol⁻¹
[tex]H_{vap}[/tex] (heat of vaporization) = 38.56 kJ⋅mol⁻¹ = 38560 J⋅mol⁻¹
㏑ [tex](\frac{760 torr}{P_{2} } )\\[/tex] = [tex]\frac{38560 J.mol^{-1} }{8.314 J.K^{-1} mol^{-1} }[/tex] [tex](\frac{1}{323.15 K } - \frac{1}{351.55 } )[/tex]
㏑ [tex](\frac{760 torr}{P_{2} } )\\[/tex] = 4638 X 2.5 X [tex]10^{-4[/tex]
㏑ [tex](\frac{760 torr}{P_{2} } )\\[/tex] = 1.159
[tex](\frac{760 torr}{P_{2} } )\\[/tex] = [tex]e^{1.159}[/tex]
[tex](\frac{760 torr}{P_{2} } )\\[/tex] = 3.188
[tex]P_{2}[/tex] = [tex]\frac{760 torr}{3.188}[/tex] = 238.3 torr
The vapor pressure of Ethanol at 15°C is calculated as 237.88torr
HOW TO CALCULATE VAPOR PRESSURE:
- The vapor pressure of ethanol in this question can be calculated using Clausius-Clapeyron Equation as follows:
- ln (P1/P2) = ∆H vap/R (1/T2 - 1/T1)
Where;
- T₁ (temperature)= (50°C + 273.15) K = 323.15K
- P₁ (vapor pressure) = ?
- T₂ (temperature) = (78.4°C + 273.15) K = 351.55K
- P₂ (vapor pressure) = 760 torr
- R = Ideal gas constant = 8.314 J⋅K ⁻¹mol⁻¹
- ∆H (heat of vaporization) = 38.56 kJ⋅mol⁻¹ = 38560 J⋅mol⁻¹
Hence;
- ln (P1/760) = 38560/8.314 (1/351.55 - 1/323.15)
- ln (P1/760) = 4637.96 (0.002845 - 0.003095)
- ln (P1/760) = 4637.96 × - 0.00025
- ln (P1/760) = -1.16
- P1/760 = exp (-1.16)
- P1/760 = 0.313
- P1 = 760 × 0.313
- P1 = 237.88torr.
- Therefore, vapor pressure of Ethanol at 15°C is calculated as 237.88torr
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