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A certain substance has a heat of vaporization of 64.08 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.50 times higher than it was at 355 K?

Respuesta :

MrRoyal

Answer:

[tex]T_2 =388.50K[/tex]

Explanation:

Given

[tex]p_1 = ??[/tex] -- Initial vapour pressure

[tex]p_2 = 6.50p_1[/tex] --- Final vapour pressure

[tex]T_1 = 355K[/tex] ---- Initial temperature

[tex]T_2 = ??[/tex] --- Final temperature

[tex]\triangle _{vap}H = 64.08kJmol[/tex] --- Enthalpy of vaporization

Required

Calculate T2

To do this, we make use of Clausius-Clapeyron equation.

Which states that:

[tex]ln(\frac{p_2}{p_1}) = \frac{\triangle _{vap}H}{R}(\frac{1}{T_1} - \frac{1}{T_2})[/tex]

Where:

[tex]R = 8.314 J.K^{-1}mol^{-1}[/tex] --- Universal Gas constant

[tex]\triangle _{vap}H = 64.08kJmol[/tex]

[tex]\triangle _{vap}H = 64080\ kJmol[/tex]

[tex]ln(\frac{p_2}{p_1}) = \frac{\triangle _{vap}H}{R}(\frac{1}{T_1} - \frac{1}{T_2})[/tex] becomes

[tex]ln(\frac{6.50p_1}{p_1}) = \frac{64080}{8.314}(\frac{1}{355} - \frac{1}{T_2})[/tex]

[tex]ln(\frac{6.50p_1}{p_1}) = 7707.48(\frac{1}{355} - \frac{1}{T_2})[/tex]

[tex]ln(6.50) = 7707.48(\frac{1}{355} - \frac{1}{T_2})[/tex]

[tex]1.872 = 7707.48(\frac{1}{355} - \frac{1}{T_2})[/tex]

Take LCM

[tex]1.872 = 7707.48(\frac{T_2 - 355}{355T_2})[/tex]

[tex]1.872 = 7707.48*\frac{T_2 - 355}{355T_2}[/tex]

[tex]1.872 = \frac{7707.48*(T_2 - 355)}{355T_2}[/tex]

Cross Multiply

[tex]355T_2*1.872 = 7707.48*(T_2 - 355)[/tex]

[tex]664.56T_2 = 7707.48T_2 - 2736155.4[/tex]

Collect Like Terms

[tex]664.56T_2 - 7707.48T_2 =- 2736155.4[/tex]

[tex]-7042.92T_2 =- 2736155.4[/tex]

Make T2 the subject

[tex]T_2 =\frac{- 2736155.4}{-7042.92}[/tex]

[tex]T_2 =388.497299416[/tex]

[tex]T_2 =388.50K[/tex]

The final temperature is 388.50K

ajeigbeibraheem

The temperature at which the vapor pressure will be 6.50 times higher than it was at 355K is 389 K

The Clausius-Clapeyron Equation is widely used for the determination of the vapor pressure of another temperature, provided we know the vapor pressure at a certain temperature as well as the heat of vaporization.

Given that:

  • the heat of vaporization [tex]\mathbf{\Delta H_{vap} = 64.08 \ kJ/mol = 64.08 \times 10^3 J/mol}[/tex]
  • the initial temperature T₁ = 355 K
  • the final temperature T₂ = ???
  • if the pressure at T₁ = P₁
  • then, the pressure at T₂ = 6.5P₁
  • the universal gas constant = 8.314 J/K/mol

By using the Clausius-Clayperon Equation:

[tex]\mathbf{In \Big( \dfrac{P_2}{P_1} \Big) = \dfrac{\Delta H_{vap}}{R} \Big (\dfrac{1}{T_1} - \dfrac{1}{T_2} \Big) }[/tex]

[tex]\mathbf{In \Big( \dfrac{6.5P_1}{P_1} \Big) = \dfrac{64.08 \times 10^3 }{8.314} \Big (\dfrac{1}{355} - \dfrac{1}{T_2} \Big) }[/tex]

[tex]\mathbf{In(6.5)=7707.48 \Big (\dfrac{1}{355} - \dfrac{1}{T_2} \Big) }[/tex]

[tex]\mathbf{1.872 =21.71 - \dfrac{7707.48}{T_2} }[/tex]

[tex]\mathbf{\dfrac{7707.48}{T_2} = 21.71- 1.872 }[/tex]

[tex]\mathbf{\dfrac{7707.48}{T_2} = 19.838 }[/tex]

[tex]\mathbf{T_2= \dfrac{7707.48}{19.838} }[/tex]

[tex]\mathbf{{T_2}=388.521 \ K }[/tex]

T₂ ≅ 389 K

Learn more about Clausius-Clayperon equation here:

https://brainly.com/question/1077674?referrer=searchResults

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