Question

A certain substance has a heat of vaporization of 72.87 kJ / mol. 72.87 kJ/mol. At what Kelvin temperature will the vapor pressure be 4.50 4.50 times higher than it was at 361 K?

Answer 1

`\large \boxed{\text{385 K}}`

Step-by-step explanation:

To answer this question, we can use the Clausius-Clapeyron equation:

`\ln \left ((p_(2))/(p_(1)) \right) = \frac{\Delta_{\text{vap}}H}{R} \left((1 )/( T_(1) ) - (1)/(T_(2)) \right)`

Data:

p₁ = p₁; T₁ = 361 K

p₂ = 4.50 T₁; T₂ = ?

R = 8.314 J·K⁻¹mol⁻¹

`\Delta_{\text{vap}}H = 72.87\text{ kJ$\cdot$mol}^(-1)`

Calculation:

`\begin{array}{rcl}\ln \left ((p_(2))/(p_(1)) \right)& = & \frac{\Delta_{\text{vap}}H}{R} \left( (1)/(T_(1)) - (1)/(T_(2)) \right)\\\\\ln \left ((4.50p_(1))/(p_(1)) \right)& = & (72870)/(8.314) \left((1 )/( 361) - (1)/(T_(2)) \right)\\\\\ln4.50 & = & 8765 \left((1 )/( 361) - (1)/(T_(2)) \right)\\\\1.504 & = & 24.28 - (8765)/(T_(2))\\\\\end{array}\\`

`\begin{array}{rcl}(8765)/(T_(2)) & = & 22.78\\\\T_(2) & = &(8765)/(22.78)\\\\& = & \textbf{385 K}\\\end{array}\\\text{The vapour pressure will be 4.50 times as high at $\large \boxed{\textbf{385 K}}$ as it was at 361 K.}`