OA02 Thermal Physics and 1D Kinematics PDF

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Mastering Physics assignment OA02 Thermal Physics and 1D Kinematics...

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OA02: Thermal Physics and 1D Kinematics Due: 11:59pm on Sunday, August 30, 2020 You will receive no credit for items you complete after the assignment is due. Grading Policy

± Volume of Copper

Part A What is the volume VV of a sample of 2.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper is 8.92 × 10 3 kg/ m 3. Express your answer in cubic centimeters to 3 significant figures.

Hint 1. How to approach the problem Since the number of moles of copper is known, calculate its mass, and then use the mass and the density of copper to find the volume of the copper. Hint 2. What is the mass of the copper? Calculate the mass mm of 2.00 mol of copper. Express your answer in grams. ANSWER: mm = 127 g

Hint 3. Using the density in calculations Recall that density is defined as the ratio of the mass to volume: ρ = M / V. Be careful with the units, since the volume should be expressed in cubic centimeters.

ANSWER: VV = 14.2 cm 3

Correct

± Air Bubble Rising in a Lake A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is 4.00 C, and the temperature at the surface is 23.0 C.

Part A What is the ratio of the volume of the bubble as it reaches the surface (V sV_s) to its volume at the bottom (V bV_b)?

Hint 1. How to approach the problem Use the ideal gas equation to calculate the ratio of the volume V sV_s at the surface to the volume V bV_b at the bottom of the lake. Be careful about the units when performing the calculations. Hint 2. Mass of air in the bubble

Since the bubble is surrounded entirely by the water, it can be assumed that no air can enter or leave the bubble during its ascent, so the number of moles nn must be a constant. Hint 3. Find an expression for the volume ratio Find a symbolic expression for the volume ratio V s / V b. Express your answer in terms of the temperature T bT_b and pressure p bp_b at the bottom of the lake and the temperature T sT_s and pressure p sp_s at the surface.

Hint 1. Using the ideal gas law Recall that the ideal gas law is pV = nRT, or V=

nRT p

,

where pp is the pressure, VV is the volume, nn is the number of moles of gas, RR is the ideal gas constant, and TT is the temperature. Set up two copies of this equation for the two situations: at the surface, where the variables are V sV_s, p sp_s, and T sT_s, and at the bottom, where they are V bV_b, p bp_b, and T bT_b. If you divide the first of these equations by the second, then you will have the ratio V s / V b of the volumes. We are assuming that nn will stay constant, because gas does not escape or enter the bubble as it rises, and RR is always a constant, so both should cancel out of your expression.

ANSWER: p sT s p bT b p bT b Vs Vb

=

p sT s p bT s p sT b p sT b p bT s

ANSWER: Vs Vb

= 3.74

Correct

Part B If Jacques were to hold his breath the air in his lungs would be kept at a constant temperature. Would it be safe for Jacques to hold his breath while ascending from the bottom of the lake to the surface? ANSWER:

yes no

Correct If Jacques were holding his breath, then air would be unable to enter or leave his lungs. As he ascends to the surface, the air in his lungs would expand, like the air in the bubble, and his lungs would have to stretch outward to hold this increased volume, which would be extremely unsafe. In fact, even if he does not hold his breath, if he ascends too quickly after a particularly long or deep dive, the nitrogen dissolved in his bloodstream could form into small bubbles, which can be equally dangerous to any diver. This condition is known as decompression sickness, or more commonly as the bends.

Prelecture Concept Question 17.06

Part A If both the pressure and volume of a given sample of an ideal gas are doubled, what happens to the temperature of the gas in Kelvins? ANSWER: The temperature of the gas is increased by two times its original value. The temperature remains constant. The temperature of the gas in increased by four times its original value. The temperature of the gas is reduced to one-fourth its original value. The temperature of the gas is reduced to one-half its original value.

Correct

PhET Tutorial: Gas Properties The behavior of gases has been studied by many scientists, and they have established certain laws known as the "gas laws." The gas laws, when combined together, give us the ideal gas equation: PV = nRT, where P is the pressure of the gas, V is the volume occupied by the gas, n is the number of moles of gas, R is the universal gas constant, and T is the temperature. Select the image below to start this simulation on the ideal gas equation.

In the Ideal tab of this simulation, you may adjust the volume, temperature, and number of molecules for a gas and observe its behavior. You can keep the pressure, the volume, or the temperature of the gas constant using the radio buttons in the menu Hold Constant to observe the wall collisions and other parameters of the gas.

Part A Charles's law states that the volume (V) of a fixed quantity of gas is directly proportional to its temperature (T) at a constant pressure. V

T

You can verify this law by plotting the graph of a gas's volume versus its temperature. To perform this analysis, set the number of "Heavy" gas molecules to 100 using the arrows to the left and right of the textbox for "Heavy" in the menu named Particles. Once the pressure reaches the value of about 11.7 atm, click on the "Pressure ↕ V" radio button under the menu Hold Constant, which is at the top right corner of the simulation. At the bottom of this menu, select "Width" to see the measurement for the width of the container in nm. Set the temperature by using the slider on the heat control below the container to add or remove heat as given in the table below. Temperature (K)

200.

250.

300.

350.

Use the width reported by the measurementbelow the chamber. Complete the table below with your raw data for the width of the box at each temperature. Drag the appropriate labels to their respective targets.

Hint 1. How to approach the problem Set the actual temperature to the given values in the table by adding or removing heat using the burner. Notice that the width of the box expands or contracts to a certain value. This will happen only if you keep the pressure constant by clicking on the "Pressure ↕ V" radio button under the menu Hold Constant. The box is lying such that the height (vertical) and length (into the page) remain constant as the width fluctuates. As the gas obeys Charles's law, the width will change in accordance with the volume required to keep the pressure constant at the temperature set. Use the measurement shown to identify the width of the box. Hint 2. Determine the width of the box at 250. K Set the temperature to 250Kand determine the width of the box when the pressure is 11.7 atm. Express your answer to two significant figures and include the appropriate units. ANSWER: 8.3 nm

ANSWER:

Reset

5.0

10.0

6.7

11.7

8.3

13.3

6.7

Correct

8.3

10.0

11.7

Help

Part B The chamber is a box with a fixed height and length which are both 4.3 nm. Using your data from Part A for the width of the chamber, plot the graph of the gas's volume versus its temperature. Temperature (K)

200.

250.

300.

350.

Width of box (nm)

6.7

8.3

10.0

11.7

Sketch a graph of the volume versus temperature.

Hint 1. How to approach the problem Use the widths of the box determined in Part A and the given height and length to calculate the volume of the box. The volume of the box is calculated using the height (h), length (l), and width (w): volumeofbox = h × l × w Make sure that all of your units are in nm when calculating the volume of the box. The volume will then have units ofnm 3. Once you have the volume measurements, plot the volume in nm 3 on the y axis (vertical), and the given temperature values in K on the x axis (horizontal). Hint 2. Calculate the volume of the box having a width of 8.3 nm The volume of the box is calculated by multiplying the height (h), length (l), and width (w) by each other. Since the height and length are the same, h = l and the equation simplifies to h 2w. Calculate the volume of the box if the width of the box is 8.3 nm, and the height and length are both 4.3 nm. Express the volume to three significant figures and include the appropriate units. ANSWER: VV = 153 nm 3

Hint 3. Determine the volumes of the box at other values of temperature Complete the table below with the volume of the box at each temperature. Drag the appropriate labels to their respective targets. ANSWER: Reset

28.8

50.3

35.7

4.3

43.0

92.5

124

153

185

216

Help

ANSWER:

No elements selected 300

250

200

150

O

200

250

300

350

400

Select the elements from the list an main menu.

Correct You can observe the graph of volume versus temperature as a straight line. Thus, you can say that V = constant × T, or V / T = constant, which can be compared to the equation for a straight line, y = mx + b. You have just demonstrated Charles's law graphically.

Part C Boyle's law states that the volume (V) of a fixed quantity of gas is inversely proportional to the gas's pressure (P) at a constant temperature: V

1 P

You can verify this law by plotting the graph of volume versus the inverse of pressure (1 /P). To perform this analysis, first set the number of "Heavy" gas molecules to 100 using the arrows to the left and right of the textbox for "Heavy" in the menu named Particles. Next, remove heat using the slider on the heat control below the container to set the temperature to 298 K. Now, select "Temperature (T)" from the menu Hold Constant, which is at the top right corner of the simulation. At the bottom of this menu, select "Width" to see the measurement for the width of the container in nm. Set the width of the box by moving the adjustable wall of the container on the left side as given in the table.

Width of box (nm)

6.0

8.0

10.0

12.0

Note the corresponding pressure reading in atm. The pressure of the container does not remain constant because the molecules exert pressure on the walls of the box. For example, if the pressure of the container varies between 11.2 and 12.0 atm, consider the average pressure of 11.6 atm. Complete the table below with your raw data for the pressure in the container at each width. Drag the appropriate labels to their respective targets.

Hint 1. How to approach the problem In the simulation, first set the initial conditions required for verifying Boyle’s law. Set the temperature to 298 K and the number of molecules of the "Heavy Species" to 100. Then, select the radio button for "Temperature (T)" from the menu Hold Constant. Now, determine the pressure in the container at each width by moving the adjustable wall of the container. You can select "Width" to see the measurement for the width of the container displayed below the box. The box is lying such that the height (vertical) and length (into the page) remain constant as the width fluctuates.Once you move the walls of the container, ensure that the gas molecules attain equilibrium. Read off the maximum and minimum pressure at this width in the Pressure gauge on the right side of the box, and record the average in atm. Hint 2. Find the pressure when the width of the box is 8.0 nm Set the width of the box to 8.0 nm and the temperature to a constant value of 298 K. Determine the pressure exerted by the gas molecules as indicated by the pressure gauge. Observe that the pressure of the box does not remain constant because the molecules exert pressure on the walls of the box. For example, if the pressure of the box varies between 11.2 and 12.0 atm, consider the pressure as 11.6 atm for calculation purposes. Express the pressure in atmosphere to one decimal place. ANSWER: 14.5 atm

ANSWER: Reset

8.4

14.5

9.8

19.3

11.6

27.1

19.3

14.5

11.6

Help

9.8

Correct

Part D The chamber is a box with a fixed height and length both of 4.3 nm. Using the widths 6.0, 8.0, 10.0, and 12.0 nm, calculate the volume of the container. Then determine the reciprocal (inverse) of the pressure (1 /P) using your data in Part C. Width of box (nm)

6.0

8.0

10.0

12.0

Pressure (atm)

19.3

14.5

11.6

9.8

Plot the graph of the volume (V) versus the reciprocal (inverse) of pressure (1 /P).

Sketch a graph of the inverse pressure versus volume.

Hint 1. How to approach the problem Use the given height and length of 4.3 nmand the given widths for the box used in Part C to find the container volume (V). The volume of the box is calculated using the height (h), length (l), and width (w): volumeofbox = h × l × w Make sure that all of your units are in nm when calculating the volume of the box. The volume will then have units ofnm 3. Note the pressure at this volume as determined in Part C, and calculate the inverse of this pressure. The inverse of the observed pressure is equivalent to 1 /P. Once you have the volume and inverse pressure calculations, plot the volume in nm 3 on the y axis (vertical), and the inverse pressure in atm − 1 on the x axis (horizontal). Hint 2. Calculate the volume of the box having a width of 8.0 nm The volume of the box is calculated by multiplying the height (h), length (l), and width (w) by each other. Since the height and length are the same, h = l and the equation simplifies to h 2w. Calculate the volume of the box if the width of the box is 8.0 nm, and the height and length are both 4.3 nm. Express the volume to three significant figures and include the appropriate units. ANSWER: VV = 148 nm 3

Hint 3. Find the inverse of 14.5 atm pressure The inverse of the pressure is equivalent to 1 /P. Find the inverse of 14.5 atm. Express your answer numerically to two significant figures. ANSWER: 1 / P = 6.9×10−2 atm − 1

Hint 4. Determine the volume of the box and the inverse of pressure for all points Complete the table below with the volume of the box and the inverse of the pressure when the width of the box is 6.0, 10.0, and 12.0 nm, respectively. Drag the appropriate labels to their respective targets. ANSWER:

Reset

0.120

259

111

185

222

0.0518

0.0862

0.102

ANSWER:

No elements selected 0.150

0.100

0.050

0.000 0

Select the elements from the list an main menu.

50

100

150

200

250

Help

Correct You can observe the graph of volume versus the inverse pressure as a straight line. Thus, you can say that V = constant × 1 / P, or PV = constant, which can be compared to the equation for a straight line, y = mx + b. You have just demonstrated Boyle's law graphically.

Combining the gas laws In addition to Charles's law and Boyle's law, there is another law known as Avogadro's law, which establishes a relationship between the number of gas molecules and the volume of a gas at a constant pressure and temperature. If V is the volume of the gas andn is the number of moles of the gas at constant pressure Pand temperature T, then according to Avogadro's law, V n. Combining Charles's law, Boyle's law, and Avogadro's law produces the ideal gas equation stated as: PV = nRT In this equation, R is known as the universal gas constant. The value for the universal gas constant R is 0.08206 L atm/ (mol K). You can see the following: PV = constant, when n and T are fixed according to Boyle's law. V / T = constant, when P and n are fixed according to Charles's law. V / n = constant, when P and T are fixed according to Avogadro's law.

Part E Calculate the volume of the gas when the pressure of the gas is 1.40 atm at a temperature of 298 K. There are 200. mol of gas in the container. The value for the universal gas constant R is 0.08206 L atm/ (mol K) . Express your answer numerically to four significant figures.

Hint 1. How to approach the problem To determine the volume of the box, use the ideal gas equation. The ideal gas equation is PV = nRT. Rearrange this equation to solve for volume, then plug the values given in the problem into this equation, and calculate the volume. Hint 2. Identify the rearranged ideal gas equation to find the volume of the gas The ideal gas equation is PV = nRT. Identify the rearranged equation to solve for V. ANSWER:

V=

nRT P

PV = nRT V = nRTP V=

P nRT

ANSWER: volume = 3493 L

Correct PV = nRT, and thus V = nRT / P. The volume of the gas is calculated by substituting the values of n, R, T, and P: V=

200.mol × 0.08206L atm / ( mol K ) × 298K 1.40atm

= 3493L

Prelecture Concept Question 17.03

Part A A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? ANSWER: The hole in the center of the washer will expand. The hole in the center will remain the same size. The hole in the center of the washer will contract. Changes in the hole cannot be determined without know the composition of the metal.

Correct

Coefficient of Volume Expansion for Water Conceptual Question The anomalous expansion characteristics of liquid water are crucial to many biological systems. Rather than an approximately constant value for the coefficient of volume expansion, the value for water changes drastically, as illustrated in the figure.

Part A Below what temperature TT does water shrink when heated? Express your answer numerically in degrees Celsius.

Hint 1. Coefficient of volume expansion

The fractional change in volume of a substance when subject to a change in temperature depends on the amount of the temperature change and the substance. The coefficient of volume expansion represents the fractional change in volume per degree of temperature change. A large coefficient of expansion means that the substance expands by a relatively large amount. Hint 2. Volume expansion relation The dependence of the change in volume of a substance (ΔVDelta V) on the initial volume (V 0V_0), the temperature change ( ΔTDelta T), and the coefficient of volume expansion (βbeta) is ΔV = βV 0ΔT Hint 3. Negative coefficient of volume expansion Since the relationship for volumetric expansion is ΔV = βV 0ΔT if βbeta is less than zero, the change in volume will be negative when the change in temperature is positive.

ANSWER: TT = 4

C

Correct

Part B If the temperature of water at 30 C is raised by 1 C, the water will expand. At approximately what initial temperature TT will water expand by twice as much when raised by 1 C? Expr...