A sample of \(5.00 \mathrm{~mL}\) of diethylether $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\( density \)=0.7134 \mathrm{~g} / \mathrm{mL}\( ) is introduced into a \)6.00-\mathrm{L}$ vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(R_{\mathrm{N}_{2}}=21.08 \mathrm{kPa}\) and \(P_{\mathrm{O}_{2}}=76.1 \mathrm{kPa}\). The temperature is held at \(35.0^{\circ} \mathrm{C},\) and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

Short Answer

Expert verified
The partial pressure of diethylether is \(20.87 \text{kPa}\) and the total pressure in the container is \(118.05 \text{kPa}\).

Step by step solution

01

Find moles of diethylether

Given that the density of diethylether is 0.7134 g/mL, and we have 5 mL of diethylether: Mass of diethylether = Density × Volume Mass of diethylether = 0.7134 g/mL × 5 mL = 3.567 g The molecular formula for diethylether is \(C_2H_5OC_2H_5\). To find its molar mass, we can sum up the atomic masses of the constituent elements: Molar mass of diethylether = 2×(Molar mass of C) + 10×(Molar mass of H) + 1×(Molar mass of O) Molar mass of diethylether = 2×(12.01 g/mol) + 10×(1.01 g/mol) + 1×(16.00 g/mol) = 74.12 g/mol Now we can find the moles of diethylether: moles of diethylether = mass of diethylether/molar mass of diethylether moles of diethylether = 3.567 g / 74.12 g/mol = 0.0481 mol
02

Calculate partial pressure of diethylether

Temperature = 35.0 °C = 308.15 K Volume of vessel = 6.00 L We will use the Ideal Gas Law to calculate the partial pressure of diethylether: \(PV = nRT\) \(P_{diethylether} = \frac{n_{diethylether} × R × T}{V}\) We will use the gas constant R = 8.314 J/(mol·K), hence we need to convert volume from liters to cubic meters: V = 6.00 L × (1 m³ / 1000 L) = 0.00600 m³ Then, we have: \(P_{diethylether} = \frac{0.0481 \text{mol} \times 8.314 \text{J/(mol·K)} \times 308.15 \text{K}}{0.00600 \text{m³}} = 20868.5 \text{Pa}\) \(P_{diethylether} = 20.87 \text{kPa}\)
03

Calculate total pressure

Now we have the partial pressures of N2, O2, and diethylether: \(P_{N_2} = 21.08 \text{kPa}\) \(P_{O_2} = 76.1 \text{kPa}\) \(P_{diethylether} = 20.87 \text{kPa}\) Total pressure in the container = Sum of partial pressures Total pressure = \(P_{N_2} + P_{O_2} + P_{diethylether}\) Total pressure = 21.08 kPa + 76.1 kPa + 20.87 kPa = 118.05 kPa So, the partial pressure of diethylether is 20.87 kPa and the total pressure in the container is 118.05 kPa.

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Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(m^{1 / 2}\). (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.

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