Answer:
[tex]\large \boxed{\text{385 K}}[/tex]
Explanation:
To answer this question, we can use the Clausius-Clapeyron equation:
[tex]\ln \left (\dfrac{p_{2}}{p_{1}} \right) = \dfrac{\Delta_{\text{vap}}H}{R} \left(\dfrac{1 }{ T_{1} } - \dfrac{1}{T_{2}} \right)[/tex]
Data:
p₁ = p₁; T₁ = 361 K
p₂ = 4.50 T₁; T₂ = ?
R = 8.314 J·K⁻¹mol⁻¹
[tex]\Delta_{\text{vap}}H = 72.87\text{ kJ$\cdot$mol}^{-1}[/tex]
Calculation:
[tex]\begin{array}{rcl}\ln \left (\dfrac{p_{2}}{p_{1}} \right)& = & \dfrac{\Delta_{\text{vap}}H}{R} \left( \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right)\\\\\ln \left (\dfrac{4.50p_{1}}{p_{1}} \right)& = & \dfrac{72870}{8.314} \left(\dfrac{1 }{ 361} - \dfrac{1}{T_{2}} \right)\\\\\ln4.50 & = & 8765 \left(\dfrac{1 }{ 361} - \dfrac{1}{T_{2}} \right)\\\\1.504 & = & 24.28 - \dfrac{8765}{T_{2}}\\\\\end{array}\\[/tex]
[tex]\begin{array}{rcl}\dfrac{8765}{T_{2}} & = & 22.78\\\\T_{2} & = &\dfrac{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.}[/tex]
