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Chapter 12: Problem 54

A \(2.50 \mathrm{g}\) sample of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is sealed in a \(5.00 \mathrm{L}\) flask at \(120.0^{\circ} \mathrm{C}\) (a) Show that the sample exists completely as vapor. (b) Estimate the temperature to which the flask must be cooled before liquid water condenses.

Short Answer

Expert verified
The sample exists completely as vapor in the given exercise scenario, and the flask must be cooled to approximately \(80^{\circ} C\) for liquid water to start condensing. This performance of calculations assumes that water behaves ideally and the ideal gas law applies.

Step by step solution

01

Converting Mass to Moles

The mass of water is given as 2.50g. To make it useful for the gas law equations, it needs to be converted into moles. This is done using the molecular weight of water, which is about 18.015 g/mol. So the number of moles (\(n\)) of water = \(2.50 \, g / 18.015 \, g /mol = 0.1387 \, mol\).
02

Realizing Water as Ideal Gas

Given that the water is sealed in the flask, operating at a high temperature of \(120.0^{\circ} C = 393.15 \, K\), the water can be approximated to behave as an ideal gas. Therefore, the ideal gas law \(PV = nRT\) (where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature) can be applied.
03

Estimating the Pressure

Using the ideal gas law, it is possible now to calculate the pressure (\(P\)) inside the flask. \(P = nRT/V = 0.1387 \, mol * 0.08206 \, L-atm/K-mol * 393.15 \, K / 5.00 \, L = 0.912 \, atm\). The pressure inside the flask is below the vapor pressure of water at \(120.0^{\circ} C\) (which is around 1.99 atm). This indicates that water exists completely as vapor in the flask. Therefore, (a) is validated.
04

Clausius-Clapeyron Equation

Next, to estimate the temperature at which the liquid water condenses, the Clausius-Clapeyron equation which describes the phase transitions can be used. The equation is \(ln(P_2/P_1) = -(\Delta H_{vap} / R) * (1/T_2 - 1/T_1)\). Here, \(P_1\) and \(P_2\) are the initial and final pressures, \(T_1\) and \(T_2\) are the corresponding temperatures, \(\Delta H_{vap}\) is the enthalpy of vaporization, and \(R\) is the gas constant.
05

Solving for Final Temperature

At \(1 atm\), water starts to condense. So, the final pressure (\(P_2\)) is 1 atm. Using \(P_2 = 1 atm\), \(P_1 = 0.912 atm\), \(\Delta H_{vap} = 40.7 KJ/mol = 40700 J/mol\), \(R = 8.314 J/mol-K\), and \(T_1 = 393.15 K\) in the Clausius-Clapeyron equation, solving for \(T_2\) gives \(T_2 =353 K\) or \(80^{\circ} C\). This means the flask must be cooled down to approximately \(80^{\circ} C\) for the water vapor to start condensing. So, the answer to (b) is \(80^{\circ} C\).

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