9/13/2015 Ch 18 HW Ch 18 HW Due: 11:59pm on Monday, September 14, 2015
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Exercise 18.4
∘ A 3.00L 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 C
Correct
Part B
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
Correct
The Ideal Gas Law Derived
The ideal gas law, discovered experimentally, is an equation of state that relates the observable state variables of the gaspressure, temperature, and density (or quantity per volume):
pV = NkBT (or pV = nRT ),
where N is the number of atoms, n is the number of moles, and R and kB are ideal gas constants such that R = NA kB, where NA is Avogadro's number. In this problem, you should use Boltzmann's constant instead of the gas constant R.
Remarkably, the pressure does not depend on the mass of the gas particles. Why don't heavier gas particles generate more pressure? This puzzle was explained by making a key assumption about the connection between the microscopic world and the macroscopic temperature T . This assumption is called the Equipartition Theorem.
The Equipartition Theorem states that the average energy associated with each degree of freedom in a system at T 1 k T k = 1.38 × 10−23 J/K absolute temperature is 2 B , where B is Boltzmann's constant. A degree of freedom is 1 mv2 m a term that appears quadratically in the energy, for instance 2 x for the kinetic energy of a gas particle of mass Loading [MathJax]/jax/output/HTMLCSS/fonts/TeX/AMS/Regular/Main.js with velocity vx along the x axis. This problem will show how the ideal gas law follows from the Equipartition Theorem. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 1/21 9/13/2015 Ch 18 HW To derive the ideal gas law, consider a single gas particle of mass m that is moving with speed vx in a container with length Lx along the x direction.
Part A
Find the magnitude of the average force ⟨Fx ⟩ in the x direction that the particle exerts on the righthand wall of the container as it bounces back and forth. Assume that collisions between the wall and particle are elastic and that the position of the container is fixed. Be careful of the sign of your answer.
Express the magnitude of the average force in terms of m, vx , and Lx .
Hint 1. How to approach the problem ⃗ From the relationship between applied force and the change in momentum per unit time, F = dp/⃗ dt, it follows that the average force in the x direction exerted by the wall on the particle is ⟨Fx ⟩ = Δpx /Δt, where Δpx is the change in the particle's momentum upon collision with the wall and Δt is the time interval between collisions with the wall.
You want to find the force exerted by the particle on the wall. This is related to the force of the wall on the particle by Newton's 3rd law.
Hint 2. Find the change in momentum
Find Δpx , the change in momentum of the gas particle when it collides elastically with the righthand wall of its container.
Express your answer in terms of m and vx .
Hint 1. Finding the final momentum
The formula for the momentum of a particle p ⃗ of mass m traveling with velocity v ⃗ is p⃗ = mv.⃗ What is the x component of the final momentum of the gas particle (i.e., after the collision)?
Express your answer in terms of m and vx . ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 2/21 9/13/2015 Ch 18 HW
pfx = −mvx
ANSWER:
Δpx = −2mvx
Hint 3. Find the time between collisions Use kinematics to find Δt, the time interval between successive collisions with the righthand wall of the container.
ANSWER:
2L Δt = x vx
ANSWER:
2 mvx ⟨Fx ⟩ = Lx
Correct
Part B
Imagine that the container from the problem introduction is now filled with N identical gas particles of mass m. The 2 particles each have different x velocities, but their average x velocity squared, denoted ⟨vx ⟩, is consistent with the Equipartition Theorem.
Find the pressure p on the righthand wall of the container. Express the pressure in terms of the absolute temperature T , the volume of the container V (where V = Lx Ly Lz), kB , and any other given quantities. The lengths of the sides of the container should not appear in your answer.
Hint 1. Pressure in terms of average force The pressure is defined as the force per unit area exerted on the wall by the gas particles. The area of the righthand wall is A = Ly Lz. Thus, if the average force exerted on the wall by the particles is ⟨Fx ⟩, then the pressure is given by
⟨Fx ⟩ p = . LyLz
Hint 2. Find the pressure in terms of velocity
Find the pressure p1 on the righthand wall due to a single particle whose squared speed in the x direction is 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 3/21 9/13/2015 1 Ch 18 HW 2 vx .
Express your answer in terms of vx , m, Lx , Ly , and Lz.
ANSWER:
2 mvx p1 = Lx LyLz
Hint 3. Find pressure in terms of temperature 2 To find the pressure from particles with average squared speed vx , you can use the Equipartition Theorem. Find the pressure p1 due to a single particle.
Express the pressure due to a single particle in terms of kB , T , Lx , Ly , Lz, and any other given quantities.
Hint 1. Relate velocity and temperature 2 Use the Equipartition Theorem to find an expression for m⟨vx ⟩.
Express your answer in terms of the gas temperature T , kB , and given quantities. ANSWER:
2 m⟨vx ⟩ = kB T
ANSWER:
kBT p1 = Lx LyLz
ANSWER:
p N k T = V B
Correct
Very good! You have just derived the ideal gas law, generally written pV = NkBT (or pV = nRT ).
This applet shows a small number of atoms in an ideal gas. On the right, the path of a specific atom is followed. Look at this for different temperatures to get a feel for how temperature affects the motions of the atoms in an ideal gas.
Part C
Which of the following statements about your derivation of the ideal gas law are true? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 4/21 9/13/2015 Ch 18 HW Check all that apply. ANSWER:
2 2 The Equipartition Theorem implies that ⟨vx ⟩ = ⟨vy ⟩. 2 2 ⟨vx ⟩ = ⟨vy ⟩ owing to inelastic collisions between the gas molecules.
With just one particle in the container, the pressure on the wall (at x = Lx) is independent of Ly and Lz.
With just one particle in the container, the average force exerted on the particle by the wall (at x = Lx) is independent of Ly and Lz.
Correct
Part D
If you heat a fixed quantity of gas, which of the following statements are true?
Check all that apply. ANSWER:
The volume will always increase. If the pressure is held constant, the volume will increase. The product of volume and pressure will increase. The density of the gas will increase. The quantity of gas will increase.
Correct
Up, Up, and Away
Hot air balloons float in the air because of the difference in density between cold and hot air. Consider a balloon in which the mass of the pilot basket together with the mass of the balloon fabric and other equipment is mb . The volume of the hot air inside the balloon is V1 and the volume of the basket, fabric, and other equipment is V2. The absolute temperature of the cold air outside the balloon is Tc and its density is ρc . The absolute temperature of the hot air at the bottom of the balloon is Th (where Th > Tc). The balloon is open at the bottom, so that the pressure inside and outside the balloon is the same here. Assume that we can treat air as an ideal gas. Use g for the magnitude of the acceleration due to gravity.
Part A
What is the density ρh of hot air inside the balloon? Assume that this density is uniform throughout the balloon.
Express the density in terms of Th , Tc, and ρc .
Hint 1. Find density in terms of temperature and pressure https://session.masteringphysics.comp/Vmy=ct/ansRsigTnmentPrintView?assignmentID=3740013 5/21 9/13/2015 Ch 18 HW Use the ideal gas law, pV = nRT , to find an expression for the density ρ of an ideal gas in terms of its temperature and pressure.
Express the density in terms of T , p, R, and m, the mass of one mole of gas.
Hint 1. Find density in terms of mass and volume
Derive an expression for the density ρ of an gas in terms of the volume occupied by the gas, V , the number of moles of gas particles, n, and the mass per mole, m. ANSWER:
ρ mn = V
ANSWER:
mp ρ = RT
Hint 2. How to use your general density equation
You now have an expression for the density of a gas in terms of temperature, mass, and pressure. Use this result to write one equation for the density of the hot air (at the bottom of the balloon) and another for the density of the cold air. Then divide the two expressions (remembering that the pressure at the bottom of the balloon and mass per mole of the hot air are the same as those of the cold air) to find a relationship between the temperatures and densities alone. Finally, use the assumption that the density is constant throughout the balloon, so the density you will have found is the density everywhere inside the balloon.
ANSWER:
\texttip{\rho _{\rm h}}{rho_h} = \large{\frac{T_{c}}{T_{h}}{\rho}_{c}}
Correct
Part B
What is the total weight \texttip{W}{W} of the balloon plus the hot air inside it?
Express your answer in terms of quantities given in the problem introduction and/or \texttip{\rho _{\rm h}} {rho_h}. ANSWER:
\texttip{W}{W} = \large{\left(m_{b}+\frac{{\rho}_{c} T_{c}}{T_{h}}V_{1}\right)g}
Correct
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 6/21 9/13/2015 Ch 18 HW
Part C
What is the magnitude of the buoyant force \texttip{F_{\rm B}}{F_B} on the balloon? Express your answer in terms of \texttip{g}{g}, \texttip{\rho _{\rm c}}{rho_c}, \texttip{V_{\rm 1}}{V_1}, and \texttip{V_{\rm 2}}{V_2}.
Hint 1. How to approach the problem
According to Archimedes' principle, the buoyant force is equal to the weight of the (cold) air displaced by the balloon and the (hot) air inside it.
Hint 2. Volume of displaced air
What is the volume of cold air displaced by the balloon and its contents? ANSWER:
Volume = V_{1}+V_{2}
ANSWER:
\texttip{F_{\rm B}}{F_B} = g{\rho}_{c}\left(V_{1}+V_{2}\right)
Correct
Part D
For the balloon to float, what is the minimum temperature \texttip{T_{\rm min}}{T_min} of the hot air inside it?
Express the minimum temperature in terms of \texttip{T_{\rm c}}{T_c}, \texttip{V_{\rm 1}}{V_1}, \texttip{V_{\rm 2}}{V_2}, \texttip{m_{\rm b}}{m_b}, and \texttip{\rho _{\rm c}}{rho_c}.
Hint 1. How to approach the problem
The balloon will just begin to float when the magnitude of the buoyant force is equal to the magnitude of the force of gravity (the balloon's weight). So, to find the minimum required temperature, set the weight of the balloon equal to the buoyant force and solve for the resulting temperature of the hot air.
Use the result of Part A to eliminate \texttip{\rho _{\rm h}}{rho_h} from your answer, if necessary.
ANSWER:
\texttip{T_{\rm min}}{T_min} \large{T_{c}\left(\frac{V_{1}}{\left(V_{1}+V_{2}\right){\frac{m_{b}} = {{\rho}_{c}}}}\right)}
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 7/21 9/13/2015 Ch 18 HW
Correct
The answer, a bit more elegantly displayed, is
\large{T_{\rm min} = T_{\rm c}\frac{V_1}{(V_1+V_2)m_{\rm b}/\rho_{\rm c}}}.
This equation implies that if the mass of the balloon alone is too large, m_{\rm b} > \rho_{\rm c} (V_1+V_2), then the balloon cannot get off the ground at any temperature.
For a given volume and balloon mass (sometimes known as the payload), the larger the balloon volume, the lower the temperature required for the balloon to float.
Mathematically, it is possible that T_{\rm min} < T_{\rm c}, but this makes little physical sense. You can show that this would only be the case if the average density of the payload is less than that of the cold air; if this were the case, no balloon would be needed!
Exercise 18.27
Part A
How many moles are there in a 1.20 {\rm kg} bottle of water?
ANSWER:
M = 66.7 {\rm mol}
Correct
Part B
How many molecules are there in the bottle?
ANSWER:
N = 4.01×1025
Correct
Equipartition Theorem and Microscopic Motion
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 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 8/21 9/13/2015 Ch 18 HW 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 \texttip{T}{T} is (1/2)k_{\rm B}T, where k_{\rm B} = 1.38 \times 10^{23}\; {\rm J/K} is Boltzmann's constant.
The average energy of the ith degree of freedom is \avg{U_i}=(1/2)k_{\rm B}T, 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)M {v_x}^2 is the kinetic energy of a gas particle of mass \texttip{M}{M} with velocity component \texttip{v_{\mit x}}{v_x} along the x axis.
The Equipartition Theorem follows from the fundamental postulate of statistical mechanicsthat 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=Nk_{\rm B}T, 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 pressure) of the gas particles' motion, only the temperature. It provides strong evidence for the validity of the Equipartition Theorem as applied to a particle gas:
\large{\frac{1}{2}M\avg{v_x^2}=\frac{1}{2}M\avg{v_y^2}=\frac{1}{2}M\avg{v_z^2}=\frac{1}{2}k_{\rm B}T} or \large{\frac{1} {2}k_{\rm s} x^2} for a particle constrained by a spring whose spring constant is \texttip{k_{\rm s}}{k_s}. If a molecule has moment of inertia \texttip{I}{I} about an axis and is rotating with angular velocity \texttip{\omega }{omega} about that axis with associated rotational kinetic energy (1/2) I \omega_x^2, that angular velocity represents another degree of freedom.
Part A
Consider a monatomic gas of particles each with mass \texttip{M}{M}. What is v_{x,{\rm{rms}}} = \sqrt{\avg{v_x^2}}, the root mean square (rms) of the x component of velocity of the gas particles if the gas is at an absolute temperature \texttip{T}{T}? Express your answer in terms of \texttip{T}{T}, \texttip{k_{\rm B}}{k_B}, \texttip{M}{M}, and other given quantities.
Hint 1. Equipartition for one velocity component
For this case, the Equipartition Theorem reduces to \large{\frac{1}{2}M\avg{v_x^2}=\frac{1}{2}k_{\rm B}T}.
ANSWER:
v_{x,{\rm{rms}}} = \sqrt{\avg{v_x^2}}= \large{\sqrt{\frac{k_{B} T}{M}}}
Correct
Part B https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 9/21 9/13/2015 Ch 18 HW Now consider the same systema monatomic gas of particles of mass \texttip{M}{M}except in three dimensions. Find v_{\rm{rms}}, the rms speed if the gas is at an absolute temperature \texttip{T}{T}.
Express your answer in terms of \texttip{T}{T}, \texttip{k_{\rm B}}{k_B}, \texttip{M}{M}, and other given quantities.
Hint 1. Equipartition Theorem for three degrees of freedom
What is the internal energy \avg{U} of a monotomic ideal gas with three degrees of freedom? Give your answer in terms of \texttip{k_{\rm B}}{k_B} and \texttip{T}{T}. ANSWER:
\large{\avg{U} = \frac{1}{2}M v_{\rm{rms}}^2 =} \large{\frac{3}{2} k_{B} T}
ANSWER:
v_{\rm{rms}}=\sqrt{\avg{v^2}}= \large{\sqrt{\frac{3k_{B} T}{M}}}
Correct
Part C
What is the rms speed \texttip{v_{\rm 0}}{v_0} of molecules in air at 0^\circ {\rm C}? Air is composed mostly of {\rm N}_2 molecules, so you may assume that it has molecules of average mass 28.0 \times 1.661 \times 10^{27}\; {\rm kg} = 4.65 \times 10^{26}\; {\rm kg}. Express your answer in meters per second, to the nearest integer.
ANSWER:
\texttip{v_{\rm 0}}{v_0} = 493 \rm m/s
Correct
Not surprisingly, this speed (a little over 760 miles per hour) is slightly higher than the speed of sound in air.
Now consider a rigid dumbbell with two masses, each of mass \texttip{m}{m}, spaced a distance \texttip{d}{d} apart.
Part D
Find \sqrt{\avg{\omega_x^2}}, the rms angular speed of the dumbbell about a single axis (taken to be the x axis), assuming that the dumbbell is lined up on the z axis and is in equilibrium at temperature \texttip{T}{T}.
Express the rms angular speed in terms of \texttip{T}{T}, \texttip{k_{\rm B}}{k_B}, \texttip{m}{m}, \texttip{d} {d}, and other given quantities.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 10/21 9/13/2015 Ch 18 HW Hint 1. Rotational energy equal to (1/2)k_{\rm B}T
What is the kinetic energy of rotation that is equal to (1/2)k_{\rm B}T by the Equipartition Theorem? Express your answer in terms of the x component of the angular velocity \texttip{\omega _{\mit x}} {omega_x} and the moment of inertia about this axis \texttip{I_{\mit x}}{I_x} ANSWER:
\large{\frac{1}{2}k_{\rm B}T} = \large{\frac{1}{2} I_{x} {{\omega}_{x}}^{2}}
Hint 2. Moment of inertia of a dumbbell
What is \texttip{I_{\mit x}}{I_x}, the moment of inertia of the dumbbell? Express \texttip{I_{\mit x}}{I_x} in terms of \texttip{m}{m} and \texttip{d}{d}.
Hint 1. Finding \texttip{I_{\mit x}}{I_x} of a dumbbell There are two atoms, each with mass m, but each is only a distance d/2 from the center of rotation (i.e., the center of mass).
ANSWER:
\texttip{I_{\mit x}}{I_x} = \large{2 m \left(\frac{d}{2}\right)^{2}}
ANSWER:
\sqrt{\avg{\omega_x^2}} = \large{\sqrt{\frac{2k_{B} T}{md^{2}}}}
Correct
Part E
What is the typical rotational frequency \texttip{f_{\rm rot}}{f_rot} for a molecule like {\rm N}_2 at room temperature (25\;^\circ \rm{C})? Assume that \texttip{d}{d} for this molecule is 1\; {\rm \AA} =10^{10}\; {\rm m}. Take the total mass of an {\rm N}_2 molecule to be m_{\rm N_2} = 4.65 \times 10^{26}\; {\rm kg}. You will need to account for rotations around two axes (not just one) to find the correct frequency.
Express \texttip{f_{\rm rot}}{f_rot} numerically in hertz, to three significant figures. ANSWER:
\texttip{f_{\rm rot}}{f_rot} = 1.34×1012 Hz
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 11/21 9/13/2015 Ch 18 HW
Correct
This frequency corresponds to light of wavelength 0.22 mm and is in the farinfrared region of the electromagnetic spectrum (which at 658 GHz is also the top end of the frequency range that can be produced with microwave techniques). However, because the molecules in the air are homonuclear diatomic molecules, their symmetry prevents them from interacting strongly with radiation of this frequency. Only nonhomonuclear molecules such as water vapor absorb energy at infrared frequencies.
Exercise 18.38
Part A
Calculate the mean free path of air molecules at a pressure of 3.50×10−13 {\rm atm} and a temperature of 291 {\rm K} . (This pressure is readily attainable in the laboratory.) Model the air molecules as spheres with a radius of 2.00×10−10 {\rm m} . ANSWER:
\lambda = 1.59×105 {\rm m}
Correct
Particle Gas Review
A particle gas consists of \texttip{N}{N} monatomic particles each of mass \texttip{m}{m} all contained in a volume \texttip{V}{V} at temperature \texttip{T}{T}. Your answers should be written in terms of the Boltzmann constant \texttip{k_{\rm B}}{k_B} and Avagadro's number \texttip{N_{\rm A}}{N_A} rather than R=N_{\rm A} k_{\rm B}.
Part A
Find \avg{v^2}, the average speed squared for each particle. Express the average speed squared in terms of the gas temperature \texttip{T}{T} and any other given quantities.
Hint 1. how to approach the problem
Since v=\sqrt{v_x^2+v_y^2+v_z^2}, it follows that \avg{v^2}=\avg{v_x^2}+\avg{v_y^2}+\avg{v_z^2} where \avg{v_i^2} is the average velocity squared for each particle in the i direction. Find the average velocity squared for each particle along each direction, then find the algebraic sum.
Hint 2. Find \avg{v_x^2} for each particle
Find \avg{v_x^2}, the average x velocity squared for each particle. Express the average x velocity squared in terms of the gas temperature \texttip{T}{T} and any other given quantities. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 12/21 9/13/2015 Ch 18 HW
Hint 1. How to approach the problem
Consider a single gas particle moving only in the x direction. Find one formula for its kinetic energy in terms of its velocity and another formula in terms of the gas temperature. Set these equations equal to each other and solve for v_x^2.
Hint 2. Kinetic energy in terms of temperature A monatomic particle has three degrees of freedom, one for translation in each of three dimensions. The kinetic energy associated with one degree of freedom (for example, motion in the x direction) is K = (1/2) k_{\rm B} T.
Hint 3. Kinetic energy in terms of velocity
Another expression for a particle's kinetic energy associated with motion in the x direction is K = (1/2) m \avg{v_x^2}.
ANSWER:
\avg{v_x^2} = \large{\frac{k_{B} T}{m}}
Hint 3. Relating the \texttip{x}{x}, \texttip{y}{y}, and \texttip{z}{z} velocities Particles of an ideal gas move randomly, so in a contained volume no direction is preferred. Therefore \avg{v_x^2}=\avg{v_y^2}=\avg{v_z^2}.
ANSWER:
\avg{v^2} = \large{\frac{3k_{B} T}{m}}
Correct
Part B
Find \texttip{U}{U}, the internal energy of the gas.
Express the internal energy in terms of the gas temperature \texttip{T}{T} and any other given quantities.
Hint 1. How to approach the problem
In an ideal gas, intermolecular forces are negligible. The internal energy of the whole gas is a sum of the kinetic energies of each gas particle.
Hint 2. Kinetic energy of a single gas particle
The kinetic energy of a single gas particle is K = (1/2)m\avg{v^2}. As stated in the problem introduction, there are a total of \texttip{N}{N} particles comprising the gas.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 13/21 9/13/2015 Ch 18 HW ANSWER:
\texttip{U}{U} = \large{{\frac{3}{2}}Nk_{B} T}
Correct
Part C Find \texttip{C_{\mit V}}{C_V}, the molar heat capacity (heat capacity per mole) of the gas at constant volume. Express the molar heat capacity in terms of \texttip{N_{\rm A}}{N_A} and \texttip{k_{\rm B}}{k_B}.
Hint 1. Definition of heat capacity
Heat capacity is generally defined to be dQ/dT. From the first law of thermodynamics,
dQ=dU+dW. At constant volume dW=0. Therefore, the heat capacity at constant volume is given by
C_V=dU/dT.
This question asks you to find the molar heat capacity, or heat capacity per mole of gas particles. The easiest way to do this is to compute heat capacity assuming that there is exactly one mole of gas (i.e., N=N_{\rm A}).
ANSWER:
\texttip{C_{\mit V}}{C_V} = \large{{\frac{3}{2}}N_{A} k_{B}}
Correct
You can use the molar heat capacity \texttip{C_{\mit V}}{C_V} to find the total heat capacity of the gas at constant volume \texttip{c_{\mit V}}{c_V}. Since \texttip{c_{\mit V}}{c_V} is the total heat capacity of the gas, it is just the heat capacity per mole multiplied by the number of moles of gas particles n=N/N_{\rm A}. Hence, c_V = (3/2)Nk_{\rm B}.
Part D
Express the pressure \texttip{p}{p} of the gas in terms of its energy density U/V. Enter the numerical factor that multiplies U/V in your expression for \texttip{p}{p} to at least three significant figures.
Hint 1. How to approach the problem Find an expression that relates the pressure \texttip{p}{p} to the volume \texttip{V}{V} of the gas. Then use your expressions for \texttip{U}{U} from Part C to write \texttip{p}{p} in terms of U/V.
Hint 2. Find an expression for pressure https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 14/21 9/13/2015 Ch 18 HW Find \texttip{p}{p}, the pressure of the gas using the Ideal Gas Law.
Express your answer in terms of \texttip{N}{N}, \texttip{T}{T}, \texttip{V}{V}, and \texttip{k_{\rm B}} {k_B}.
Hint 1. Ideal gas law
The ideal gas law is pV=N k_{\rm B} T.
ANSWER:
\texttip{p}{p} = \large{\frac{N k_{B} T}{V}}
ANSWER:
\texttip{p}{p} = 0.667 U/V
Correct
As an interesting aside, note that for an ultrarelativistic particle gas, such as a gas of photons, \large{p = \frac{1}{3}\frac{U}{V}}.
Now imagine that the mass of each gas particle is increased by a factor of 3. All other information given in the problem introduction remains the same.
Part E
What will be the ratio of the new molar mass M^\prime to the old molar mass \texttip{M}{M}?
ANSWER:
\large{\frac{M^\prime}{M}} = 3
Correct
Part F
What will be the ratio of the new rms speed v_{\rm rms}^\prime to the old rms speed \texttip{v_{\rm rms}}{v_rms}?
Hint 1. Definition of rms speed
By definition v_{\rm rms}\equiv\sqrt{\avg{v^2}}. A monatomic particle has three degrees of freedom, one for translation in each of three dimensions. The kinetic energy associated with one degree of freedom (for example, motion in the x direction) is K = (1/2) k_{\rm B} T. Another expression for a particle's kinetic energy associated with motion in the x direction is K = (1/2) m \avg{v_x^2}.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 15/21 9/13/2015 Ch 18 HW ANSWER:
\large{\frac{v_{\rm rms}^\prime}{v_{\rm rms}}} = 0.577
Correct
So if the internal energy (temperature) remains the same, when the mass of the particles increases, the velocity correspondingly decreases. Observe that the pressure and so the average force that the particles exert on the walls also remains the same.
Part G What will be the ratio of the new molar heat capacity C_V^\prime to the old molar heat capacity \texttip{C_{\mit V}} {C_V}?
Hint 1. How to approach the problem The total heat capacity of the gas can be determined by multiplying the heat capacity per mole by the number of moles of gas particles n=N/N_{\rm A}. How does the heat capacity per mole depend on the mass of the particles? How does the number of moles of gas particles depend on the mass of the particles?
ANSWER:
\large{\frac{C_V^\prime}{C_V}} = 1
Correct
In fact, the molar heat capacity is (almost) the same for all monoatomic gases, as the general expression derived earlier shows. For example, helium, neon, and argon all have almost the same molar heat capacity. The tiny differences in the values tell us that the ideal gas equation describes gases pretty well, but it is not perfect.
± Gas Scaling
When doing numerical calculations involving temperature, you need to pay particular attention to the temperature scale you are using. In general, you should use the Kelvin scale (for which T=0 represents absolute zero) in such calculations. This is because the standard thermodynamic equations (i.e., the ideal gas law and the formula for energy of a gas in terms of temperature) assume that zero degrees represents absolute zero.
If you are given temperatures measured in units other than kelvins, convert them to kelvins before plugging them into these equations. (You may then want to convert back into the initial temperature unit to give your answer.)
Part A
The average kinetic energy of the molecules of an ideal gas at 10 ^\circ {\rm C} has the value \texttip{K_{\rm 10}} {K_10}. At what temperature \texttip{T_{\rm 1}}{T_1} (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K_{10}? Express the temperature to the nearest integer. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 16/21 9/13/2015 Ch 18 HW
Hint 1. Formula for energy in terms of temperature From the Equipartition Theorem we know that each translational degree of freedom of particles in a gas contributes (1/2)k_BT to the average internal energy, where the temperature \texttip{T}{T} is measured in kelvins.
Recalling that a gas molecule is free to move in three perpendicular directions, give an expression for the kinetic energy \texttip{K_{\rm tr}}{K_tr} of a molecule due to its translational degrees of freedom. Express your answer in terms of \texttip{k_{\rm B}}{k_B} and \texttip{T}{T}.
ANSWER:
\texttip{K_{\rm tr}}{K_tr} = \large{\frac{3}{2} k_{B} T}
Hint 2. Convert from Celsius to Kelvin scale
Don't forget to convert the degrees Celsius into kelvins before plugging in for the temperature: T_{\rm Celsius} = T_{\rm Kelvin} 273.
ANSWER:
\texttip{T_{\rm 1}}{T_1} = 293 ^\circ {\rm C}
Correct
Part B
The molecules in an ideal gas at 10 ^\circ {\rm C} have a rootmeansquare (rms) speed \texttip{v_{\rm rms}} {v_rms}. At what temperature \texttip{T_{\rm 2}}{T_2} (in degrees Celsius) will the molecules have twice the rms speed, 2v_{\rm rms}? Express the temperature to the nearest integer.
Hint 1. What is rms speed? The rootmeansquare or rms speed of molecules in a gas is equal to the square root of the average velocity squared of the molecules: v_{\rm rms}=\sqrt{(v^2)_{\rm av}}. The rms speed is related to the translational kinetic energy \texttip{K_{\rm tr}}{K_tr} of the gas via the following equation:
\large{\frac{1}{2}mv^2_{\rm rms}=K_{\rm tr}},
where \texttip{m}{m} is the mass of the gas molecules.
Hint 2. Find the change in translational kinetic energy
Assume that, at 10^\circ {\rm C}, the molecules have translational kinetic energy \texttip{K_{\rm 1}}{K_1}. When the rms speed of the molecules is doubled, what is their new translational kinetic energy \texttip{K_{\rm 2}}{K_2}? Express your answer in terms of \texttip{K_{\rm 1}}{K_1}. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 17/21 9/13/2015 Ch 18 HW ANSWER:
\texttip{K_{\rm 2}}{K_2} = 4 K_{1}
ANSWER:
\texttip{T_{\rm 2}}{T_2} = 859 ^\circ {\rm C}
Correct
Exercise 18.43
Part A
Compute the specific heat capacity at constant volume of nitrogen ({\rm N}_{2}) gas. The molar mass of {\rm N}_{2} is 28.0 {\rm g}/{\rm mol}. ANSWER:
c = 741 {\rm J/(kg \cdot K)}
Correct
Part B
You warm 1.65 {\rm kg} of water at a constant volume from 15.0 {\rm ^\circ C} to 28.5 {\rm ^\circ C} in a kettle. For the same amount of heat, how many kilograms of 15.0 {\rm ^\circ C} air would you be able to warm to 28.5 {\rm ^\circ C}? Make the simplifying assumption that air is 100\% {\rm N}_{2}. ANSWER:
m = 9.32 {\rm kg}
Correct
Part C
What volume would this air occupy at 15.0 {\rm ^\circ C} and a pressure of 1.03 {\rm atm} ? ANSWER:
V = 7640 {\rm L} https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 18/21 9/13/2015 Ch 18 HW
Correct
Exercise 18.52
A physics lecture room has a volume of 255 {\rm m^3} .
Part A
For a pressure of 1.00 \rm atm and a temperature of 27.0 {\rm ^\circ C}, use the idealgas law to estimate the number of air molecules in the room. Assume all the air is \rm N_{2}.
Express your answer to three significant figures and include the appropriate units. ANSWER:
N = 6.24×1027 \rm molecules
Correct
Part B
Calculate the particle densitythat is, the number of \rm N_{2} molecules per cubic centimeter. Express your answer to three significant figures and include the appropriate units. ANSWER:
particle density = 2.45×1019 \rm molecules/{\rm cm^3}
Correct
Part C Calculate the mass of the air in the room. Express your answer to three significant figures and include the appropriate units.
ANSWER:
m = 290 {\rm kg}
Correct
Problem 18.68 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 19/21 9/13/2015 Ch 18 HW
Part A The size (radius) of an oxygen molecule is about 2.0 \times 10^{10}\; {\rm{ m}}. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from idealgas behavior at ordinary temperatures (T= 300{\rm{ K}} ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself. Express your answer using one significant figure. ANSWER:
P = 1×108 \rm Pa
Correct
Problem 18.76
The surface of the sun has a temperature of about 5800 {\rm K} and consists largely of hydrogen atoms.
Part A Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67 \times 10^{ 27} {\rm kg}.) ANSWER:
V = 1.20×104 {\rm m/s}
Correct
Part B
The escape speed for a particle to leave the gravitational influence of the sun is given by (2GM/R)^{1/2}, where M is the sun's mass, R its radius, and G the gravitational constant. The sun`s mass is M = 1.99\times 10^{30}{\rm kg}, its radius R = 6.96 \times 10^{8}{\rm m} and G=6.673 \times 10^{11} {\rm N \cdot m^{2} /kg^{2}}. Calculate the escape speed for the sun. ANSWER:
v = 6.18×105 {\rm m/s}
Correct
Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3740013 20/21 9/13/2015 Ch 18 HW Can appreciable quantities of hydrogen escape from the sun? ANSWER:
Yes No
Correct
Part D
Can any hydrogen escape? ANSWER:
Yes No
Correct
Score Summary: Your score on this assignment is 101%. You received 12.15 out of a possible total of 12 points.
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