Gas Properties and Balloons & Buoyancy SI M Homework Answer K ey

Transcription

1 Gas Properties and Balloons & Buoyancy SI M Homework Answer K ey 1) In class, we have been discussing how gases behave and how we observe this behavior in our daily lives. In this homework assignment, you will need to use the Gas Properties Simulation to help you develop a visual and conceptual model of how the bulk properties of a gas (such as pressure, temperature and volume) relate to what the individual gas molecules are doing and how this enters into your everyday life. a) Put a moderate amount of gas into a constant volume box. Without varying any of the controls, what general information about the nature of gases can you extract from the ribe the visual picture you have of a gas, noting any characteristic features of gases that you observe from the applet. You should include at least 3 things. Think of things that make a gas unique and distinguishable from a solid.) The atoms (or molecules) of gas are all bouncing around randomly and independently. Unlike in a solid, they are not bound together in any way and are free to move around and change places relative to each other. The density is much lower than it would be in a solid. Also, the density changes on small scales as small groups of atoms randomly come together and then separate again. The gas spreads out to fill all the space available to it, unlike a solid object which has a constant volume. b) Look at the animation of the particles bouncing around in the volume. Describe what visual information you can use to get a sense of the pressure that the gas particles are exerting on the walls. Play around with controls and see how the pressure responds. What visual cues are associated with an increase in pressure? Be sure to focus not only on the walls, but also on the gas in the center of the box, what visual cues can you use to get a sense of the pressure of the gas at any point in the volume? The pressure exerted on the wall should be directly related to the number of particles hitting the wall per unit time and how fast they are moving when they hit. If the particles are moving faster (hitting the wall harder and more often), or if there were a higher particle density (more particles to hit the wall), we would interpret these changes as signs of increased pressure. For the gas in the center of the box, even though there is no wall present, the pressure depends on the same factors. Higher particle density and higher speeds correspond to higher pressure. c) It is well known that the higher you climb in the mountains, the less oxygen there is (fewer air molecules). (Use the simulation to create a gravitational attraction to the floor). i) Describe the changes you observe in the distribution of air particles within the atmosphere as you increase the gravitational attraction.

2 With gravity turned on, the particle density increases toward the bottom of the box and decreases near the top. As the strength of the gravity increases, the difference in density between the lower and upper portions of the box becomes greater. ii) What do the visual changes you observed in the distribution of the gas particles tell you about how the pressure and temperature of the gas varies throughout the box? How is this consistent or inconsistent with you experience in real life? The density of the gas is highest at the bottom of the box and decreases smoothly with increasing height. Since the gas density is directly proportional to the pressure, we can conclude that the pressure is also highest at the bottom of the box and decreases with height. This is consistent with the fact that in real life, the pressure and density of air is lower at higher altitudes. For example, people coming to Boulder from sea level tend to get out of breath easily (indicating lower density of air, including oxygen) and any sealed containers brought here from sea level tend where the container was sealed). iii) Explain the physics behind WHY the observation that there is less oxygen the higher you climb makes sense in terms of the forces acting in the system (namely gravity and air pressure). If we consider any small volume of air somewhere in space, the net force on that volume of air is zero otherwise it would be accelerating in some direction. We know that the small volume of air experiences a downward force from gravity equal to the weight of the air. It also experiences pressure (force per unit area) on all sides from the surrounding air. So the upward pressure on the bottom on this volume must be slightly greater than the downward pressure on the top of the volume, so the net force exerted by the surrounding air is upward, balancing the downward force of gravity. Now if we decrease the gravitational force, the pressure must also decrease since the pressure and the gravitational forces balance each other. The force of gravity is slightly weaker at higher elevations, which means that the air pressure at higher elevations is lower than at lower elevations. According to the ideal gas law P*V = k * (# air particles) *T, the number of air molecules (within a given volume at a fixed temperature) must decrease with a decrease in pressure. So at higher elevations there is less oxygen. d) Using the visual model of the behavior of gas molecules presented in the simulation explain how a suction cup works. The suction cup is being

3 pulled towards the wall. pushed towards the wall. neither pushed nor pulled. Explain what is happening with the forces that cause the cup to stay against the wall and that support your answer whether the suction cup is being pulled, pushed, or something else? When a suction cup is pressed onto a wall, the air is pushed out. Then the suction cup tries to return to its natural shape, but if the smooth outer edge of the sucti let the air back in. So there is very little air inside the suction cup, and the pressure is lower inside than outside. Since the pressure is higher on the outside, the air molecules on the outside bombard the suction cup and push it against the wall. Using your understanding of the forces, why does the suction cup start to fail if it leaks too much? If the suction cup leaks too much, then enough air gets in that the pressure on the inside approaches the pressure on the outside, and the difference is not enough to hold it against the wall. 2) In reading the car manual you see that they tell you to check the tire pressure when the tires are cold, and that it should read 24 psi with the tire gauge (so 24 psi + 14 psi atmospheric pressure gives 38 total psi inside the tire). Unfortunately you have read this only after having driven on them for 100 miles and you have measured that the gauge reads 27 psi. You have 1000 miles more to drive and so it is important to know if they are under-inflated or over-inflated. Fortunately you have taken Physics 1020, so you take out a thermometer and measure that the tire temperature is about 30 Celsius HOTTER than the 25 C air temperature. Given that your current tire pressure read 27 psi according to the gauge, what would the gauge read for pressure if you allowed the tires to cool to be the same as the air temperature of 25 C (which is the temperature at which you were supposed to check the tires)? According to the ideal gas law, P*V = k * (# air particles) *T, pressure and temperature are directly proportional. Since the volume of the tire and number of air molecules in the tire are fixed, we can rearrange the equation so that all of the things that are changing are on the left side and all of the things that remain constant are on the right side. P/T = k * (# air particles)/ V

4 Before driving on the tires, the ideal gas law tells us that the pressure P1 and temperature T1 of the tires are related by the equation P1/T1 = k * (# air particles)/ V. After driving on the tires, the new tire pressure P2 and the new tire temperature T2 are related by a similar equation P2/T2 = k * (# air particles)/ V. Since the right hand side of both equations remains constant, we can set the left hand sides equal to each other P1/T1=P2/T2 Solving for P1, we find that P1=P2*T1/T2. The total pressure in the tire is P2=(27+14) psi = 41 psi and the temperatures of the tire before and after driving are T1=25 C=298 K and T2=(25+30) C=328 K respectively. Therefore the pressure of the cold tire is P1=(41 psi)*(298/328)=37.25 psi. After subtracting of the atmospheric pressure, the gauge would measure (37.25 psi 14 psi) = psi. Explain the physics principles and reasoning you used to calculate this answer. (Again, the simulation can help you figure out how to solve this or check your reasoning. Remember, the absolute temperature and pressure readings in the simulation are not meaningful, but the proportional changes are.) If we heat up the tire, the air molecules inside the tire will move around faster and will collide into the sides of the tire with more force. This means that the air pressure inside the tire will increase at the temperature increases. According to the ideal gas law, temperature and pressure are directly proportional. Since the volume of the tire and the number of molecules in the tire remain constant, we know that ratio of the temperature and the pressure must also remain constant. We can therefore set the ratios of the pressures and temperatures for the cold and heated tires equal to each other to determine the cold tire pressure, as was done in the previous problem. 3) was well sealed before you left Denver. You land in Boston and proceed to your hotel. a) The number of air molecules within the shampoo bottle: has decreased has stayed the same has increased b) If the walls of the shampoo bottle are strong and rigid so that the bottle retains the same shape as it was before you left, how does the pressure of the air inside the bottle compare to the pressure of the air in Boulder and to the pressure of the air in Boston?

5 pressure of the air in Boston less than, less than less than, same as less than, greater than same as, less than same as, same as same as, greater than greater than, less than greater than, same as greater than, greater than c) The air pressure at sea level is 14.7 lbs/in2. The air pressure in Boulder is 20% less then the air pressure at sea level. How many lbs of force are being exerted on each square inch of the shampoo bottle? A pressure of 14.7 lbs/in^2 is exerted on the outside of the bottle, while the ambient air pressure in Boulder is exerted on the inside. The net force per square inch is the difference between that of the two locations, and since the pressure in Boulder is 20% less than that at sea level, that net force is 20% of the force per square inch at sea level, or 0.2 x 14.7 lbs = 2.94 lbs. of force on each square inch. A slightly longer but conceptually easier method is to say that the pressure in Boulder is 80% of the pressure at sea level, or 0.8 x 14.7 lbs/in^2 = lbs/in^2. This is the pressure exerted on the inside. So, the difference in force per square inch between the outside and the inside is = 2.96 lbs/in^2. d) If your shampoo bottle is like most, and its walls are not very rigid, you notice that the walls have collapsed a bit (for this problem, consider the bottle to be like a plastic Ziploc, so not at all rigid). i) pressure of the air in decreases, decreases decreases, stays the same decreases, increases stays the same, decreases stays the same, stays the same stays the same, increases increases, decreases increases, stays the same increases, increases Explain the physics principles and reasoning behind your answer. In the simulation, keep temperature constant and try changing the volume. (Note: the temperature of the gas will increase as you change the volume. Wait for the simulation to automatically return the temperature to its original value (removing it) and then evaluate how the pressure has changed and why).

6 The collapse means that the volume of the bottle has decreased a bit. Since temperature is constant, this means that the pressure inside has increased and is now greater than the air pressure in Boulder. ii) If the walls are flaccid as in the plastic Ziploc case, how will the pressure inside compare to the pressure outside the bottle when you are in Boston but have not yet opened the shampoo? the pressure inside will be lower the pressure outside will be lower the pressure inside and outside will be the same. Explain your reasoning behind your answer. If the walls are really flaccid, they will move in response to any net force on. So the bottle will collapse until the pressure on the inside is equal to the atmospheric pressure on the outside. iii) Assuming the temperature in Boston and Boulder are the same and using an original volume of 1/2 liter of air (1 L bottle). What is the new volume of air in the bottle? Using the ideal gas law, in Boulder we have: P1 = k (#)T/V1 And in Boston, P2 = k(#)t/v2. We know that k, (# particles), and T are the same for both. So we can arrange the equations with everything that might change on one side: P1 x V1 = k(#)t, and P2 x V2 = k(#)t. The right sides of these equations are the same, which means the left sides have to be equal as well. So P1 x V1 = P2 x V2. Also remember that the pressure in Boulder is 80% of the pressure in Boston, so we can express this relation as P1 = 0.8 x P2. This means (0.8 x P2) x V1 = P2 x V2, so 0.8 x V1 = V2.

7 We have V1 = 0.5 L (only the air will be compressed by the pressure difference;; the shampoo is not subject to the ideal gas law), so V2 = 0.8 x 0.5 L = 0.4 L. So, the new volume of air in the bottle in Boston is 0.4 L. iv) If you now heat your shampoo bottle up, you can get it to expand back to its original volume. Explain why this works. If we heat up the bottle, the air molecules inside gain kinetic energy and hit the sides of the bottle harder with each collision. As the temperature increases, they will begin to exert more pressure on the inside of the bottle. The walls of the bottle will respond to the pressure difference and expand until the pressure is equalized. This also makes sense if we look at the ideal gas law equation: k, and (# particles) remain the same, the weak walls of the bottle ensure the pressure will remain the same, so if we increase T, then V increases as well. 4) There are 2 balloons in a room. They are identical in size and material. Balloon #1 is filled with air and balloon #2 is filled with Helium. a) them change. For the 2 balloons in this room, the pressure of the air inside balloo less than, less than less than, same as less than, greater than same as, less than same as, same as same as, greater than greater than, less than greater than, same as greater than, greater than b) How does the number of air molecules in balloon #1 compare to the number of He atoms in balloon #2? There are more air molecules in #1 than He atoms in #2. There are fewer air molecules in #1 than He atoms in #2. The number of air molecules in #1 is the same as He atoms in #2. c) Use the Gas Properties applet to simulate the air molecules (the heavy or blue species in the simulation) inside the balloon and repeat the simulation for the same conditions (T, P, V) for Helium atoms (the light or red species in the simulation). What is different about the motion of the molecules inside the air balloon compared to the motion of the atoms inside the He balloon that supports your answers for a) and b)? Include your reasoning. At the same temperature, pressure, and volume, the helium atoms move faster

8 than the air molecules. Since they are moving faster, the helium atoms hit the walls of the balloon more often and at higher speeds than the air molecules. So even though the helium atoms have less mass, they can exert the same pressure as the same number of air molecules. d) If the volume of the balloon is 1 meter in diameter and the density of air in Boulder is 1 kg/m^3, what is the buoyancy force of the air in the room on each of the balloons? (Remember the volume of a sphere is (4* *r^3)/3 The buoyancy force is the net upward force from the differences in atmospheric pressure at different heights. The net force due to pressure differences on any small volume of air is just the right amount to hold it up against the force of gravity. Any object of the same volume experiences the same buoyancy force. For a balloon of diameter 1 m, the radius is 0.5 m so the volume of the balloon is (4/3) x 0.5^3 = 0.52 m^3. The mass of air that would occupy this volume is 0.52 kg, so the buoyancy force on the balloon is equal and opposite to the gravitational force on this mass of air: 0.52 kg x 9.8 m/s^2 = 5.1 N. e) Explain why the He balloon rises and the Air balloon falls. The gravitational force on balloon #2 plus the helium inside it is less than the upward buoyancy force on the balloon, so the helium balloon rises. On the other hand, the mass of balloon #1 plus the air inside it is enough that the gravitational force is greater than the buoyancy force, so the balloon falls. f) If the mass of the material the balloon is made of is 0.01 kg, how much mass could you attach to the He balloon and still have it float (in kg)? (The density of He gas at the pressure and temperature in the room is 0.14 kg/m^3). From above, the volume of the balloon is 0.52 m^3. So the mass of helium inside it is (0.52 m^3) x (0.14 kg/m^3) = kg. The total mass of the balloon and the helium is = kg. The gravitational force on this mass is mg = kg x 9.8 m/s^2 = 0.81 N. The upward buoyancy force is 5.1 N, so there is = 4.3 N left available to support extra weight. The mass that can be supported is m = F/g = (4.3 N)/(9.8 m/s^2) = 0.44 kg. 5) Now let s explore how hot air balloons work using the Balloons & Buoyancy Simulation. We know that hot air balloons heat the air inside them in order to rise. Observe closely what happens when you heat the air inside the balloon and when you let the air inside the balloon cool back down. a) When the heat is on and the balloon is floating, how do the number of air molecules inside the balloon compare to the number of air molecules that would be occupying that volume if hrink until

9 they are the same? The number of air molecules in the balloon is less than the number in an equivalent volume outside the balloon. The balloon is not sealed, so the pressure inside and outside are always the same. The hot air inside is at the same pressure as the surrounding atmosphere even though the density is lower because the molecules are moving at a higher average speed, so each molecule undergoes more collisions and exerts a greater force with each collision. b) Assume we have a hot air balloon with the same volume as the balloons in problem 4. i) What is the buoyancy force upwards due to the air surrounding the balloon? (Hint: look at what you did in problem 5). The buoyancy force is F_buoy = volume x (density of air displaced) x g. This hot air balloon has the same volume as the balloon in the previous problem, so the buoyancy force is the same: 5.1 N. ii) If we heat the air inside the balloon from 293 K (or 20 C) to 393 K (or 120 C), how does the mass of the air inside the balloon compare to the mass of an equivalent volume of air outside the balloon? The mass of the air inside the balloon is larger. The mass of the air inside the balloon is smaller. The mass of the air inside the balloon is the same as the mass of the air in an equivalent volume outside the balloon. If we heat the air inside the balloon from 293 K (or 20 C) to 393 K (or 120 C), what is the numerical value for the mass (in kg) of the air inside the balloon for this hotter temperature? (Hint, think about how the number of air molecules inside the balloon has changed. Is it 1/2 (or 1/3 or etc) as many molecules? ) Using the ideal gas law, we know that the initial and final pressures are the same, so k x (initial # particles) x T_initial / V = k x (final # particles) T_final / V. Also, k and V are the same, so they cancel: (initial # particles) x T_inital = (final # particles) x T_final So, (final # particles) = (initial # particles) x T_initial/T_final = (initial # particles) x 293/393 = (initial # particles) x

10 The mass is directly proportional to the number of particles. The initial mass is the same as in the air balloon in the earlier problem: 0.52 kg. So the final mass is (0.52 kg) x = kg. iii) If again the balloon material had a mass of 0.01 kg, how much mass could you attach to the hot air balloon and still have it float? The gravitational force on the balloon and the hot air inside is F = mg = (0.01 kg kg) x 9.8 m/s^2= 3.9 N. So the difference between the buoyancy force and the gravitational force is = 1.23 N, which means the balloon can carry extra mass equal to (1.23 N)/(9.8 m/s^2) = 0.12 kg. Discuss how this compares to the mass that the He balloon could lift and why it makes sense. In the previous problem, we determined that the He balloon could carry a mass of.44 kg, and that the mass of the He inside the balloon was.07 kg. The He balloon can carry more mass than the hot air balloon, which makes sense because both balloons have the same buoyant force, but the mass of the He inside the balloon is less than the mass of the air inside. Since the mass of the He balloon is lighter than the mass of the hot air balloon, the He balloon can carry more additional weight.