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Problem Set 20 – Kinetic Molecular Theory of Gases Chem 105 1. (a) What is meant by the root-mean-square (RMS) speed of gas particles? The root-mean-square (RMS) is the speed of a particle in a gas that has the average kinetic energy of all the particles of the sample. (b) Suppose you are at a track where 7 joggers are doing laps. Their speeds are (in mph) 5, 5, 4, 3, 7, 8, and 9 mph. Calculate their average speed, then calculate their RMS speed (use 3 sig figs). Average speed- 5 + 5 + 4 + 3 + 7 + 8 + 9 7 = 5.86 mph RMS is the square root of the average of the squared speeds of all the particles in of a gas √ 5 2 + 5 2 + 4 2 + 3 2 + 7 2 + 8 2 + 9 2 7 = 6.20 mph 2. (a) Does pressure affect the root-mean-square speed of the particles in a gas? Why or why not? u rms = √ 3 RT M , Pressure does not appear in the equation, therefore pressure has no effect (b) How does the RMS speed of a N 2 molecule in a gas sample change under the following conditions (assuming the gas behaves ideally): a) Increasing the temperature b) Increasing the volume c) Adding Ar gas to the container while maintaining the same T. (a) Increasing the temperature increases the rms velocity of molecules, but it is not a linear proportionality. Rms velocity squared is linearly proportional to the temperature. (b) Increasing the volume while holding the temperature constants does not affect the rms velocity (c) Mixing gases together at the same temperature does not change rms velocity because the RMS velocity only depends on T and the molar mass of the gas in question. Inter- molecular forces are negligible for ideal gases. (c) A sample of O 2 gas is compressed by transferring it to a smaller container while keeping the T constant. What change do you expect in a) the average kinetic energy of the molecules, The average kinetic energy only depends on the T of the gas, so maintaining a constant T ensures that the average kinetic energy does not change b) the average speed of the molecules, The RMS speed of the gas only depends on the molar mass of the molecule and the temperature of the gas. Because T is constant, compressing the gas has no effect on the average speed c) the number of collisions that the molecules make with the container walls per unit time.

Because the molecules are moving with the same average speed, but they are now in a smaller volume, they are expected to collide with the walls of the container more frequently. d) the pressure of the container P is inversely proportional to V. Therefore, P should increase 3. (a) Rank the gases NO, NO 2 , N 2 O 4 , and N 2 O 5 in order of increasing root-mean-square speed at 0°C. u rms = √ 3 RT M According to the equation, the lower the molar mass, the greater the root mean square speed. The molar masses of NO, NO 2 , N 2 O 4 , and N 2 O 5 are 30, 46, 92, 108 g/mol. Therefore, the rank order in terms of increasing root mean square speed is N 2 O 5 < N 2 O 4 < NO 2 < NO (b) Below is the distribution of molecular speeds of CO 2 and SO 2 molecules at 25°C. Which curve is the profile for SO 2 ? Which of the profiles should match that of propane (C 3 H 8 ), a common fuel in portable grills? Explain your reasoning. The larger the molar mass, the lower the rms speed. Curve 1 shows more molecules at lower speed and curve 2 represents fewer molecules at higher speed. The molar masses of CO 2 , SO 2 and C 3 H 8 are 44, 64, and 44 g/mol. Curve 1 represents SO 2 with higher molar mass. Curve 2 represents both CO 2 and C 3 H 8 with same molar masses. 4. (a) Gases A and B have molecular weights of 65.2 g/mol and 101.2 g/mol, respectively. What is the rms velocity of these molecules at 23 ºC? (just for fun, you can convert their speeds to mph to see if these molecules would get a ticket on the freeway) Use the equation for RMS speed: (remember to use M in kg/mol) For A u RMS = √ 3 RT M = √ 3 ( 8.314 J mol K ) ( 296 K ) 0.0652 Kg mol = 337 m / s For B u RMS = √ 3 RT M = √ 3 ( 8.314 J mol K ) ( 296 K ) 0.1012 Kg mol = 270 m / s

(b) Suppose you have a container of gas with a molecular weight of 16.48 g/mol. If the RMS speed is 852.4 m/s, what is the temperature of the gas? T= u rms 2 M/3R = 852.4 2 X 0.01648 3 X 8.314 = 480 K (c) Enriching Uranium The two isotopes of uranium, 238 U and 235 U, can be separated by diffusion of the corresponding UF 6 gases. What is the ratio of the root-mean-square speed of 238 UF 6 to that of 235 UF 6 at constant temperature? u RMS , ( U F 6 238 ) u RMS , ( U F 6 235 ) = √ M ( U F 6 235 ) M ( U F 6 238 ) = √ 349 352 = 0.996 5. (a) How is the rate of effusion of a gas related to its: (a) molar mass; (b) root-mean-square speed; (c) temperature? a) rate of effusion if directly proportional to root mean square speed which is given by, u RMS = √ 3 RT M As the molar mass increases, u RMS decreases because of the inverse relationship, and the rate of effusion will decrease b ¿ With the increase in u RMS , the rate of effusion will also increase because of the direct relationship c ¿ As, the temperature increases, u RMS increases resulting in increase of rate of effusion. (b) Which of the two outcomes shown below more accurately illustrates the effusion of gases from a balloon at constant atmospheric pressure if the red spheres have a greater root-mean-square speed than the blue spheres? Explain your reasoning. The red spheres effuse at a faster rate than blue spheres because of higher root mean square speed, therefore loss of red spheres should occur at a greater speed. Therefore, balloon a) is the correct illustration as it loses 4 red spheres as opposed to b) which loses just 1 6. (a) At a given temperature, gas A diffuses twice as fast as gas B. What is the relationship between the molar masses of the two gases? u RMS A u RMS B = r A r B = √ M B M A =2 Squaring both sides

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M B M A = 4 In other words, the molar mass of B is four times greater than that of A. (b) An unknown gas diffuses 3.728 times slower than hydrogen gas (H 2 ) at any given temperature. What is the molecular weight of the unknown gas? We can use Graham’s law of effusion/diffusion: rate H 2 rat e x = √ M x M H 2 = 3.728 3.728 ¿ 2 × ( 2.016 g mol ) = 28.01 g / mol 3.728 ¿ 2 ×M H 2 = ¿ M x = ¿ 7. Compounds sensitive to oxygen are often manipulated in glove boxes that may contain an atmosphere of pure nitrogen or pure argon. A rubber balloon filled with carbon monoxide was placed in such a glove box. After 24 hours, the volume of the balloon was unchanged. Did the glove box contain N 2 or Ar? N 2 The molar masses of the gases are CO-28g/mol, N 2 -28, Ar-40, Because CO and N 2 have same molar masses, they would effuse at the same rate and the volume won’t change 8. (a) Describe the difference between force and pressure. Force is the product of the mass of an object and that object’s acceleration. Pressure is the force exerted over a given area. F=mass*acceleration P=Force/Area (b) Why is it easier to travel over deep snow when wearing the snowshoes shown to the right rather than just boots? As , P=Force/Area, the snowboots would distribute force due to mass of the person over a large surface area, reducing the pressure.This allows walking on the top of the snow without sinking. (c) Why does an ice sakter exert more pressure on ice when wearing newly sharpened skates than when wearing skates with dull blades? The sharp blades would exert force over smaller area increasing the pressure. 9. Convert the following pressures to the indicated units: a) 694 torr in Pa 694 torr . 133.32 Pa 1 torr = 92526 Pa b) 3.5 atm in torr 3.5 atm. 760 torr 1 atm = 2660 torr c) 5.0  10 6 Pa in atm

5.0  10 6 Pa * 9.87 X 10 − 6 atm 1 Pa = 49.35 atm d) An official NCAA basketball is inflated to a pressure of about 11 psi above the normal 15 psi of atmospheric pressure. Express the total pressure in (a) atm, (b) torr, (c) Pascals The total pressure is 11 + 15 psi = 26 psi (1 atm/14.7 psi) = 1.8 atm (760 torr/14.7 psi) = 1.3  10 3 torr (1.01325  10 5 /14.7 psi) = 1.8  10 5 Pa e) Record High Atmospheric Pressure The highest atmospheric pressure ever recorded on Earth was 108.6 kPa at Tosontsengel, Mongolia, on December 19, 2001. Express this pressure in (a) millimeters of mercury, (b) atmospheres, and (c) millibars. a) 108.6 kPa x 1 atm 101.325 kPa x 760 mm Hg 1 atm = 814.6 mmHg b) 108.6 kPa x 1 atm 101.325 kPa = 1.072 atm c) 108.6 kPa x 10 mbar 1 kPa = 1086 mbar f) Record Low Atmospheric Pressure Despite the destruction from Hurricane Katrina in August, 2005, the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered an atmospheric pressure of 88.2 kPa on October 19, 2005, about 2 kPa lower than Hurricane Katrina. What was the difference in pressure between the two hurricanes in (a) millimeters of mercury, (b) atmospheres, and (c) millibars? a) 2 kPa x 1 atm 101.325 kPa x 760 mm Hg 1 atm = 15 mm Hg b) 2 kPa x 1 atm 101.325 kPa = 0.020 atm c) 2 kPa x 10 mbar 1 kPa = 20 mbar

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Related Questions

Which of the following gases would have the slowest molecular speed at 25 oC?(a) He (b) Ne (c) Ar (d) Xe (e) they are all Noble gases so their speed is the same

Compute the root-mean-square speed of H, molecules in a sample of hydrogen gas at a temperature of 76°C. |ms-

Calculate the most probable speed of oxygen (O2) molecules in a gas at T = 334 K, O2 atomic mass is 32 amu, where the atomic mass unit (amu) is 1.66 x 10-27 kg. Provide your answer in units of meters per second, but do not include the units in your answer, just the number in normal form to 3 significant digits

1.15(a) A gas at 250 K and 15 atm has a molar volume 12 per cent smaller than that calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces?

Calculate the mean speed of (a) He atoms. (b) CH4 moleculesat (i) 79 K. (ii) 315 K. (iii) 1500 K.

1.4(a) A sample of 255 mg of neon occupies 3.00 dm at 122 K. Use the perfect gas law to calculate the pressure of the gas.

Calculate the mean speed of atoms of argon in a gas at 25 oC if the mean speed of neon under these conditions is 560 m s-1. B) What is the ratio of kinetic energies of argon and neon atoms at this temperature ?

5B Use the distribution function for molecular speeds to derive an expression for the average of the square of the speed, . Compare your result to root mean square speed derived from the Kinetic Theory of Gases.

23b

1.7(a) In an attempt to determine an accurate value of the gas constant, R, a student heated a container of volume 20.000 dm³ filled with 0.251 32 g of helium gas to 500°C and measured the pressure as 206.402 cm of water in a manometer at 25°C. Calculate the value of R from these data. (The density of water at 25°C is 0.997 07 g cm3; the construction of a manometer is described in Exercise 1.6a.)

1.3 mole of helium gas at a temperature of 276 K is confined to a cubical container whose sides are 12 cm long.Find the mean summed kinetic energy of ALL the atoms in the container in [J].

At 298K, N2 molecules have a mean speed of 475 m/s. As a consequence, the probability to randomly pick 1 molecule at 475.00 m/s is extremely high. Select one: True False

Q4.2 A container, of volume 0.85 m³, has 500 g of an ideal gas at a temperature of 30°C and a pressure of 1.0 atm. (a) What is the number of moles of the gas? (b) What is the molar mass of the gas (in grams)? (c) what is element the container filled with? (check the periodic table) (d) What is the RMS speed of the gas molecules? (e) What is the average kinetic energy per molecule of the gas? (Hint: 1 atm=1.01 × 105 pa)

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