Answer:
[tex]\boxed{\text{139 $\, ^{\circ}$C}}[/tex]
Explanation:
The question is asking, "At what temperature does the vapour pressure of water equal 3.4 atm?"
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 = 1 atm; T₁ = 373.15C
p₂ = 3.4atm; T₂ = ?
R = 8.314 J·K⁻¹mol⁻¹
[tex]\Delta_{\text{vap}}H = \text{39.67 kJ//mol}[/tex]
(The enthalpy of vaporization changes with temperature. Your value may differ from the one I chose.)
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{3.4}{1} \right)& = & \dfrac{39670}{8.314} \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\\ln3.4 & = & 4771 \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\1.224 & = & 12.78 - \dfrac{4771}{T_{2}}\\\\\dfrac{4771}{T_{2}} & = & 11.56\\\\\end{array}[/tex]
[tex]T_{2} & = & \dfrac{4771}{11.56} = \text{412.6 K} = \textbf{139 $\, ^{\circ}$C}\\\\\text{The maximum temperature is } \boxed{\textbf{139 $\, ^{\circ}$C}}[/tex]
