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GASES Question 1: 1994 B Free Response Sherrie Park AP Chemistry Gangluff, Per. 3/4.

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Presentation on theme: "GASES Question 1: 1994 B Free Response Sherrie Park AP Chemistry Gangluff, Per. 3/4."— Presentation transcript:

1 GASES Question 1: 1994 B Free Response Sherrie Park AP Chemistry Gangluff, Per. 3/4

2 A student collected a sample of hydrogen gas by the displacement of water as shown by the diagram above. The relevant data are given in the following table.

3 To complete this question, you must use the equation PV= nRT correctly. You are already given the following from the table: P (Pressure)= 745 mm Hg – 23.8 mm Hg = 721.2mm Hg ** The equilibrium vapor pressure must be subtracted from atmospheric pressure to find the pressure of hydrogen gas collected. Temperature (T) = 25 C = 298 K **Must be converted into Kelvins (absolute temperature) Given these values, you can now calculate the moles of hydrogen by substituting the numbers into the equation and solving for (n). V (volume)= 90.0 mL x (1 L/ 1000mL) = 0.09 L ** must be in Liters Gas constant (R) = 62.4 L · Torr · K-1 · mol-1 **This constant is specifically used because the pressure is calculated in Torrs. A). Calculate the number of moles of hydrogen gas collected. PV=nRT (( mmHg)(0.09 L) = (x)(62.4 L · Torr · K- 1 · mol-1) (298 K) = 3.49 x 10^-3 mol H 2

4 B). Calculate the number of molecules of water vapor in the sample of gas. In order to answer this question, start by using PV=nRT to find the moles of water vapor. Then convert moles into molecules (or atoms) by multiplying the number of moles received with Avogradro’s number: x 10^23. P = 23.8mmHg V = 90.0ml/1000ml = 0.09 L N = x mol of H 2 O R = 62.4 L · Torr · K-1 · mol-1 T = 25 C = 298 K PV = nRT (23.8mmHg)(0.09 L) = (x)(62.4 L · Torr · K-1 · mol- 1)(298 K) =1.15x10^-4 moL H 2 O (1.15x10^-4 mol H 2 O)(6.022 x 10^23) = 6.93 x 10^19 molecules of H 2 O

5 C). Calculate the ratio of the average speed of the hydrogen molecules to the average speed of the water vapor molecules in the sample. The average kinetic energies of H 2 and H 2 O will be equal to one another. Therefore… (Mass H 2 )(Velocity H 2 )^2 = (Mass H 2 O )(Velocity H 2 O )^2 Equation for Average Kinetic Energy: KE = (1/2)mv^2 The masses are the molar masses of each molecule: 2 g H 2 and 18 g H 2 O 2(V H 2 )^2 = 18(V H 2 O )^2 V H 2 /V H 2 O = Square root (18/2) = 3

6 D). Which of the two gases, H 2 or H 2 0, deviates more from ideal behavior? Explain your answer. In order to answer this question, you need to know the deviations from ideal gas law behavior. Deviations are greater if: 1). Intermolecular forces of gas molecules are greater. 2). Mass (and subsequently volume) of gas molecules is greater. Basically, the larger the molecule is, the more it deviates from ideal behavior. In this case, H 2 O deviates more from ideal behavior. The is because the volume of the H 2 O molecule is greater than the volume of the H 2 molecule. It is also the larger molecule. Additionally, the intermolecular forces among the H 2 O molecules are stronger than those among the H 2 molecules. Since it has a great number of electrons, it has a greater Van Der Waals Attraction.


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