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4 September, 03:46

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?

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  1. 4 September, 04:04
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    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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