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A certain substance has a heat of vaporization of 67.49 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 291 K?

Respuesta :

znk

Answer:

309 K

Step-by-step explanation:

We can use the Clausius-Clapeyron equation to solve this problem:

ln(p₂/p₁) = (ΔHvap/R)(1/T₁- 1/T₂)

p₁ = p₁;    T₁ = 291 K

p₂ = 5p₁; T₂ = ?

R  = 8.314 J·K⁻¹mol⁻¹

ln(5p₁/p₁) = (67 490/8.314)(1/291 – 1/T₂)

        ln5 = 8118(1/291 – 1/T₂)        Remove parentheses

    1.609 = 8118/291 - 8118/T₂

    1.609 = 27.90 – 8118/T₂         Subtract 27.90 from each side

  -26.29 = -8118/T₂                      Multiply each side by -T₂

26.29T₂ = 8118                             Divide each side by 26.29

         T₂ = 8118/26.29  

        T₂ = 309 K

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