University Physics 2 - Tutorial 18 Solns PDF

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University Physics 2 - Tutorial 18 Solns...

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CHAPTER(18( ( Introduction to the Ideal Gas Law Description: Practice using the ideal gas law with a series of questions in which all but two gas parameters are held fixed. Learning Goal: To understand the ideal gas law and be able to apply it to a wide variety of situations. The absolute temperature T, volume V, and pressure p of a gas sample are related by the ideal gas law, which states that pV=nRT. Here n is the number of moles in the gas sample and R is a gas constant that applies to all gases. This empirical law describes gases well only if they are sufficiently dilute and at a sufficiently high temperature that they are not on the verge of condensing. In applying the ideal gas law, p must be the absolute pressure, measured with respect to vacuum and not with respect to atmospheric pressure, and T must be the absolute temperature, measured in kelvins (that is, with respect to absolute zero, defined throughout this tutorial as −273∘C). If p is in pascals and V is in cubic meters, use R=8.3145J/(mol⋅K). If p is in atmospheres and V is in liters, use R=0.08206L⋅atm/(mol⋅K) instead.

Part A A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0∘C. What is the new absolute pressure p2? Express your answer in terms of p1.

Hint 1. How to approach the problem To find the final pressure, you must first determine which quantities in the ideal gas law remain constant in the given situation. Note that R is always a constant. Determine which of the other four quantities are constant for the process described in this part. Check all that apply. ANSWER:

p V n T Now manipulate the ideal gas law (pV=nRT) so that n, R, and V, the constants in this situation, are isolated on the right side of the equation: PT=nRV.

Since the right side of the equation is a constant in this situation, the quantity P/T, which is always equal to nR/V, must be the same at the beginning and the end of the process. Therefore, set p1/T1=p2/T2. Plug in the values given in this part and then solve for p2, the final pressure.

Hint 2. Convert temperatures to kelvins To apply the ideal gas law, all temperatures must be in absolute units (i.e., in kelvins). What is the initial temperature T1 in kelvins? ANSWER:

T1 =

0 20 100 273 293 K The Celsius and Kelvin temperature scales have the same unit size, so to convert from degrees Celsius to kelvins, just add 273. ANSWER:

p2 = This modest temperature increase (in absolute terms) leads to a pressure increase of just a few percent. Note that it is critical for the temperatures to be converted to absolute units. If you had used Celsius temperatures, you would have predicted that the pressure should double, which is far greater than the actual increase.

Part B Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is added, the bag inflates to accommodate it. The pressure of the gas within the bag remains at 1.00 atm and its temperature remains at room temperature (20.0∘C). How many moles n have been introduced into the bag by the time its volume reaches 22.4 L? Express your answer in moles.

Hint 1. How to approach the problem Rearrange the ideal gas law to isolate n. Be sure to use the value for R in units that are consistent with the rest of the problem and hence will cancel out to leave moles at the end. ANSWER:

n = 0.932

mol

One mole of gas occupies 22.4 L at STP (standard temperature and pressure: 0 ∘C and 1 atm). This fact may be worth memorizing. In this problem, the temperature is slightly higher than STP, so the gas expands and 22.4 L can be filled by slightly less than 1 mol of gas.

Part C Some hydrogen gas is enclosed within a chamber being held at 200∘C with a volume of 0.0250 m3. The chamber is fitted with a movable piston. Initially, the pressure in the gas is 1.50×106Pa (14.8 atm). The piston is slowly extracted until the pressure in the gas falls to 0.950×106Pa. What is the final volume V2 of the container? Assume that no gas escapes and that the temperature remains at 200∘C. Enter your answer numerically in cubic meters.

Hint 1. How to approach the problem To find the final volume, you must first determine which quantities in the ideal gas law remain constant in the given situation. Note that R is always a constant. Determine which of the other four quantities are constant for the process described in this part. Check all that apply. ANSWER:

p V n T Now look at the ideal gas law: pV=nRT. Since n, R, and T are all constants in this situation, the quantity pV, which is always equal to nRT, must be the same at the beginning and the end of the process. Therefore, set p1V1=p2V2. Plug in the values given in this part and then solve for V2, the final volume. ANSWER: −2 V2 = 3.95×10

m3

Notice how n is not needed to answer this problem and neither is T, although you do make use of the fact that T is a constant.

Part D Some hydrogen gas is enclosed within a chamber being held at 200∘C whose volume is 0.0250 m3. Initially, the pressure in the gas is 1.50×106Pa (14.8 atm). The chamber is removed from the heat source and allowed to cool until the pressure in the gas falls to 0.950×106Pa. At what temperature T2 does this occur? Enter your answer in degrees Celsius.

Hint 1. How to approach the problem To find the final temperature, you must first determine which quantities in the ideal gas law remain constant in the given situation. Note that R is always a constant. Determine which of the other four quantities are constant for the process described in this part.

Check all that apply. ANSWER:

p V n T Now manipulate the ideal gas law (pV=nRT) so that n, R, and V, the constants in this situation, are isolated on the right side of the equation: PT=nRV. Since the right side of the equation is a constant in this part, the quantity P/T, which is always equal to nR/V, must be the same at the beginning and the end of the process. Therefore, set p1/T1=p2/T2. Plug in the values given in this part and then solve for T2, the final temperature. ANSWER:

T2 = 26.6

∘C

This final temperature happens to be close to room temperature. Hydrogen remains a gas to temperatures well below that, but if this question had been about water vapor, for example, the gas would have condensed to liquid water at 100∘C and the ideal gas law would no longer have applied.

( Exercise 18.4 Description: A 3.00-L tank contains air at 3.00 atm and 20.0 degree(s) C. The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. (b) If the temperature... A 3.00-L tank contains air at 3.00 atm and 20.0 ∘C. The tank is sealed and cooled until the pressure is 1.00 atm.

Part A What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. ANSWER:

T = -175 Part B

∘C

If the temperature is kept at the value found in part A and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm? ANSWER:

V = 1.00

L

Solution( 18.4.II D EN TI FY :

pV = nRT.

S ET U P : E X EC U TE :

(a) n, R and V are constant.

n, R and T are constant so

(b)

Exercise 18.16 Description: Three moles of an ideal gas are in a rigid cubical box with sides of length L. (a) What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0 degree(s) C? (b) What is the force when the... Three moles of an ideal gas are in a rigid cubical box with sides of length 0.360 m .

Part A What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C? Express your answer to three significant figures and include the appropriate units. ANSWER:

F=

= 2.03×104

Also accepted:

= 2.03×10

4

,

= 2.03×10

4

Part B What is the force when the temperature of the gas is increased to 100.0∘C? Express your answer to three significant figures and include the appropriate units. ANSWER:

F=

= 2.58×104

Also accepted:

= 2.59×10

4

,

= 2.58×10

4

Solution( 18.16.

I D EN TI FY : and pV = nRT. S ET U P : For a cube, E X EC U TE : (a) The force of any side of the cube is volume is For so

since the ratio of area to

(b) For

so

F =

nRT (3 mol)(8.314 J/mol ⋅ K)(373. 15 K) = = 3.10 × 104 N. L 0. 300 m

Exercise 18.26 Description: Consider 1 mol an ideal gas at T and P pressure. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube. (a) What is the length of an edge of ... Consider 1 mol an ideal gas at 26 ∘C and 1.10 atm pressure. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube.

Part A What is the length of an edge of each cube if adjacent cubes touch but do not overlap? ANSWER:

l=

m −9

= 3.33×10 Part B

How does this distance compare with the diameter of a typical molecule? The diameter of a typical molecule is about 10−10m. Express your answer using one significant figure. ANSWER:

ldmolecule =

= 30 Part C How does their separation compare with the spacing of atoms in solids, which typically are about 0.3 nm apart? Express your answer using one significant figure. ANSWER:

llsolid

= = 10

Solution

18.26.

I D EN TI FY :

Use

and

with

to calculate the volume V occupied by 1 molecule. The

length l of the side of the cube with volume V is given by S ET U P :

The diameter of a typical molecule is about E X EC U TE :

(a)

and

gives

(b) The distance in part (a) is about 10 times the diameter of a typical molecule. (c) The spacing is about 10 times the spacing of atoms in solids.

Exercise 18.35 Description: (a) At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at T? (Hint: The molar mass of hydrogen atoms is 1.008 g/mol and of nytrogen atoms is 14.007 g/mol. The molar mass...

Part A At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at 25.0 ∘C? (Hint: The molar mass of hydrogen atoms is 1.008 g/mol and of nytrogen atoms is 14.007 g/mol. The molar mass of H2 is twice the molar mass of hydrogen atoms, and similarly for N2.) ANSWER:

T=

= 3860

∘C

Solution 18.35.

I D EN TI FY and S ET U P :

Use equal

to relate T and M for the two gases. vrms = 3RT/M , so

where T must be in kelvins. Same

so same

for the two gases and

TN 2 /MN 2 = TH 2 /MH 2. E X EC U TE :

⎛ MN ⎞ ⎛ 28.014 g/mol ⎞ 3 2 TN 2 = TH 2 ⎜ ⎟ = [(20 + 273)K]⎜ ⎟ = 4. 071× 10 K ⎜ MH ⎟ 2 016 g/mol . ⎝ ⎠ ⎝ 2 ⎠

TN 2 =(4071 − 273)°C = 3800°C.

Problem 18.52 Description: During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature at the surface was 27.0 degree(s) C and at the bottom it was 7.0 degree(s) C. The... During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature at the surface was 27.0 ∘C and at the bottom it was 7.0 ∘C. The density of seawater is 1030 kg/m3.

Part A

A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) ANSWER:

h = 2.04 m Part B At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it? ANSWER:

p = 7.37×105 Pa

Solution 18.52.

I D EN TI FY :

Apply

to the air inside the diving bell. The pressure p at depth y below the surface of

the water is S ET U P : at the surface and at the depth of 13.0 m. E X EC U TE : (a) The height of the air column in the diving bell at this depth will be proportional to the volume, and hence inversely proportional to the pressure and proportional to the Kelvin temperature:

(1. 013× 105 Pa)

⎛ 280. 15 K⎞ ⎜ ⎟ = 0. 26 m. (1 .013 ×10 Pa) + (1030 kg/m )(9. 80 m/s )(73. 0 m)⎝ 300. 15 K ⎠ The height of the water inside the diving bell is hʹ = (2. 30 m)

5

3

(b) The necessary gauge pressure is the term

2

from the above calculation,

Problem 18.58 Description: A vertical cylindrical tank contains n of an ideal gas under a pressure of P at 20^circC. The round part of the tank has a radius of 10 cm, and the gas is supporting a piston that can move up and down in the cylinder without friction... A vertical cylindrical tank contains 1.60 mol of an ideal gas under a pressure of 0.200 atm at 20∘C. The round part of the tank has a radius of 10 cm, and the gas is supporting a piston that can move up and down in the cylinder without friction. There is a vacuum above the piston.

Part A What is the mass of this piston? Express your answer with the appropriate units. ANSWER:

m=

= 64.9 Also accepted:

= 64.8

,

= 64.9

,

= 64.9

Part B How tall is the column of gas that is supporting the piston? Express your answer with the appropriate units. ANSWER:

h=

= 6.12

Also accepted:

= 6.14

,

= 6.13

,

= 6.12

Solution 18.58.

I D EN TI FY : The upward force exerted by the gas on the piston must equal the piston’s weight. Use to calculate the volume of the gas, and from this the height of the column of gas in the cylinder. S ET U P :

with

and

For the cylinder,

(b) V = πr2h and V = nRT/p. Combining these equations gives h = nRT/πr2p, which gives

Problem 18.59 Description: A large tank of water has a hose connected to it, as shown in the figure . The tank is sealed at the top and has compressed air between the water surface and the top. When the water height h has the value h_1, the absolute pressure p of the compressed ... A large tank of water has a hose connected to it, as shown in the

figure . The tank is sealed at the top and has compressed air between the water surface and the top. When the water height h has the value 3.50 m , the absolute pressure p of the compressed air is 4.20 ×105Pa. Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be 1.00 ×105Pa.

Part A

What is the speed with which water flows out of the hose when h = 3.50 m ? ANSWER:

v=

m/s

= 26.2 Part B As water flows out of the tank, h decreases. Calculate the speed of flow for h= 3.00 m . ANSWER:

v=

m/s

= 16.1 Part C Calculate the speed of flow for h =2.10 m . ANSWER:

v=

m/s

= 6.53 Part D At what value of h does the flow stop? ANSWER:

h = 1.74

m

Solution 18.59.

I D EN TI FY : Apply Bernoulli’s equation to relate the efflux speed of water out the hose to the height of water in the tank and the pressure of the air above the water in the tank. Use the ideal-gas equation to relate the volume of the air in the tank to the pressure of the air. S ET U P : Points 1 and 2 are shown in Figure 18.59.

large tank implies

Figure 18.59 E X EC U TE : (a)

v2 = 26 .2 m/s. (b) The volume of the air in the tank increases so its pressure decreases. is the pressure for

so

and p is the pressure for

⎛ 4 .00 m − h0 ⎞ ⎛ 4 .00 m − 3 .50 m ⎞ 5 5 p = p 0⎜ ⎟ = (4. 20× 10 Pa)⎜ ⎟ = 2. 10× 10 Pa. h ⎝ 4. 00 m − ⎠ ⎝ 4. 00 m − 3. 00 m⎠ Repeat the calculation of part (a), but now

and

v2 = 5 .44 m/s. (c)

means

This is

so

with h in meters.

and

(

)

quadratic formula: h =1 15 20 . ± (15 .20)2 −4(23 .39) = (7 .60 ± 5.86) m 2 h must be less than 4.00 m, so the only acceptable value is h = 7 .60 m − 5 .86 m = 1 .74 m.

( Equipartition Theorem and Microscopic Motion Description: Discussion of Equipartition Theorem and illustration of its implications for the mechanical motion of small objects. Learning Goal: To understand the Equipartition Theorem and its implications for the mechanical motion of small objects. In statistical mechanics, heat is the random motion of the microscopic world. The average kinetic or potential energy of each degree of freedom of the microscopic world therefore depends on the temperature. If heat is added, molecules increase their translational and rotational speeds, and the atoms constituting the molecules vibrate with larger amplitude about their equilibrium positions. It is a fact of

nature that the energy of each degree of freedom is determined solely by the temperature. The Equipartition Theorem states this quantitatively: The average energy associated with each degree of freedom in a system at absolute temperature T is (1/2)kBT, where kB=1.38×10−23J/K is Boltzmann's constant. The average energy of the ith degree of freedom is ⟨Ui⟩=(1/2)kBT, where the angle brackets represent "average" or "mean" values of the enclosed variable. A "degree of freedom" corresponds to any dynamical variable that appears quadratically in the energy. For instance, (1/2)Mvx2 is the kinetic energy of a gas particle of mass M with velocity component vx along the x axis. The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically accessible quantum state of a system has equal probability of being populated, which in turn leads to the Boltzmann distribution for a system in thermal equilibrium. From the standpoint of an introductory physics course, equipartition is best regarded as a principle that is justified by observation. In this problem we first investigate the particle model of an ideal gas. An ideal gas has no interactions among its particles, and so its internal energy is entirely "random" kinetic energy. If we consider the gas as a system, its internal energy is analogous to the energy stored in a spring. If one end of the gas container is fitted with a sliding piston, the pressure of the gas on the piston can do useful work. In fact, the empirically discovered ideal gas law, pV=NkBT, enables us to calculate this pressure. This rule of nature is remarkable in that the value of the mass does not affect the energy (or the p...