If at $71226 Pa$ $"71226 Pa"$ of pressure, ethoxyethane boils at $25^{\circ} C$ $25^@ "C"$ , then what vapor pressure is needed to boil at $78^{\circ} C$ $78^@ "C"$ if across the temperature range we can assume that $DeltaH_"vap" = "29.1 kJ/mol"$ ?

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If at $71226 Pa$ $"71226 Pa"$ of pressure, ethoxyethane boils at $25^{\circ} C$ $25^@ "C"$ , then what vapor pressure is needed to boil at $78^{\circ} C$ $78^@ "C"$ if across the temperature range we can assume that $DeltaH_"vap" = "29.1 kJ/mol"$ ?

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Answer 1 · 2017-03-04T03:22:14

$"418897 Pa"$ , or $"4.13 atm"$ .

Recall the Clausius-Clapeyron equation (at least the name):

$ln(P_2/P_1) = -(DeltaH_"vap")/(R)[1/T_2 - 1/T_1]$

where:

$P$ is the vapor pressure of the liquid at the particular temperature $T$ .

$DeltaH_"vap"$ is the enthalpy for the vaporization process.

$R = "8.314472 J/mol"cdot"K"$ is the universal gas constant.

Basically, this equation describes how to determine vapor pressure at a new temperature. You were given the following info:

$DeltaH_"vap" = "29.1 kJ/mol"$

$P_1 = "71226 Pa"$ at $T_1 = 25.0^@ "C"$

$P_2 = ???$ at $T_2 = 78.0^@" C"$

The $DeltaH_"vap"$ was written as $DeltaH_"vap"^@$ because $T_1$ happened to be the same temperature as defined for standard thermodynamic temperature and pressure ( $25^@ "C"$ and $"1 atm"$ , or sometimes $"1 bar"$ in some newer texts).

Anyways, to calculate the new vapor pressure, we arbitrarily chose $P_1$ for the given pressure and are to calculate $P_2$ .

$P_2/P_1 = "exp"[-(DeltaH_"vap")/(R)[1/T_2 - 1/T_1]]$

$P_2 = P_1"exp"[-(DeltaH_"vap")/(R)[1/T_2 - 1/T_1]]$

where $"exp"("stuff")$ is the function $e^("stuff")$ to keep the equation from looking too small.

So, plug stuff in to get:

$color(blue)(P_2) = ("71226 Pa")"exp"[-(29.1 cancel"kJ/mol")/(0.008314472 cancel("kJ/mol")cdotcancel"K")[1/(78.0 + 273.15 cancel"K") - 1/(25.0 + 273.15 cancel"K")]]$

$=$ $color(blue)("418897 Pa")$

That means the vapor pressure increased at a higher temperature.

This should make sense because a higher temperature implies a higher average kinetic energy, so the molecules at the surface of the solution can escape the solution more easily, increasing the vapor pressure above the solution.

In $"torr"$ this would be $"3141.99 torr"$ , which is about $"4.13 atm"$ , so this is rather unusual. But it make sense because the boiling point of ethoxyethane is about $34^@ "C"$ , and we're well past that.

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If at $71226 Pa$ $"71226 Pa"$ of pressure, ethoxyethane boils at $25^{\circ} C$ $25^@ "C"$ , then what vapor pressure is needed to boil at $78^{\circ} C$ $78^@ "C"$ if across the temperature range we can assume that $DeltaH_"vap" = "29.1 kJ/mol"$ ?

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If at "71226 Pa"71226 Pa"71226 Pa" of pressure, ethoxyethane boils at 25^@ "C"25∘C25^@ "C", then what vapor pressure is needed to boil at 78^@ "C"78∘C78^@ "C" if across the temperature range we can assume that DeltaH_"vap" = "29.1 kJ/mol"DeltaH_"vap" = "29.1 kJ/mol"?

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If at $71226 Pa$ $"71226 Pa"$ of pressure, ethoxyethane boils at $25^{\circ} C$ $25^@ "C"$ , then what vapor pressure is needed to boil at $78^{\circ} C$ $78^@ "C"$ if across the temperature range we can assume that $DeltaH_"vap" = "29.1 kJ/mol"$ ?