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?

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