Question 1
A sample of ideal gas is taken through the thermodynamic cycle shown above. Process C is isothermal.
(a) Consider the portion of the cycle that takes the gas from state 1 to state 3 by processes A and B. Calculate the magnitude of the following and indicate the sign of any nonzero quantities.
- The net change in internal energy ΔU of the gas
- The net work W done on the gas
- The net energy Q transferred to the gas by heating
(b) Consider isothermal process C.
i. Compare the magnitude and sign of the work W done on the gas in process C to the magnitude and sign of the work in the portion of the cycle in part (a). Support your answer using features of the graph.
ii. Explain how the microscopic behavior of the gas particles and changes in the size of the container affect interactions on the microscopic level and produce the observed pressure difference between the beginning and end of process C.
(c) Consider two samples of the gas, each with the same number of gas particles. Sample 2 is in state 2 shown in the graph, and sample 3 is in state 3 shown in the graph. The samples are put into thermal contact, as shown above. Indicate the direction, if any, of energy transfer between the samples. Support your answer using macroscopic thermodynamic principles.
▶️Answer/Explanation
(a) The change in internal energy is zero because the initial and final temperatures are the same at points 1 and 3. The work done on the gas is -P∆V = -300 J. Because the work is negative, 300 J of energy must be transferred to the gas by heating in order for the internal energy of the gas to remain constant.
(b)(i) The magnitude of the work done is less than the work in part (a) because there is less area under the curve. The work is also the opposite sign from part (a) because the volume decreases, as shown by the direction of the arrow.
(ii) Temperature does not change, so the speed of the molecules and the force of collisions with the walls of the container stays the same. Volume decreases, so the density of the gas molecules increases, and they collide more frequently. This means more net force due to collisions with the container walls. The smaller volume also means less surface area.
(c) The temperature of the gas in sample 2 is higher than the temperature of sample 3. Energy goes from hot to cold, so energy will transfer from sample 2 to sample 3.
Question 2
A group of students design an experiment to investigate the relationship between the density and pressure of a sample of gas at a constant temperature. The gas may or may not be ideal. They will create a graph of density as a function of pressure. They have the following materials and equipment.
- A sample of the gas of known mass Mg in a sealed, clear, cylindrical container, as shown
- above, with a movable piston of known mass mp
- A collection of objects each of known mass mo
- A meterstick
(a)
i. Describe the measurements the students should take and a procedure they could use to collect the data needed to create the graph. Specifically indicate how the students could keep the temperature constant. Include enough detail that another student could follow the procedure and obtain similar data.
ii. Determine an expression for the absolute pressure of the gas in terms of measured quantities, given quantities, and physical constants, as appropriate. Define any symbols used that are not already defined.
iii. Determine an expression for the density of the gas in terms of measured quantities, given quantities, and physical constants, as appropriate. Define any symbols used that are not already defined.
iv. The graph above represents the students’ data. Does the data indicate that the gas is ideal? Describe the application of physics principles in an analysis of the graph that can be used to arrive at your answer.
Another group of students propose that the relationship between density and pressure could also be obtained by filling a balloon with the gas and submerging it to increasing depths in a deep pool of water.
(b) Why could submerging the balloon to increasing depths be useful for determining the relationship between the density and pressure of the gas?
(c) The balloon is kept underwater in the deep pool by a student pushing down on the balloon, as shown above. Let Vb represent the volume of the inflated balloon, mb represent the mass of just the balloon (not including the mass of the gas), pg represent the density of the gas in the balloon, and pw represent the density of the water. Derive an expression for the force the student must exert to hold the balloon at rest under the water, in terms of the quantities given in this part and physical constants, as appropriate.
▶️Answer/Explanation
(a) (i) Place the container in an ice bath, so the part below the piston is submerged. Measure the radius and height of the piston. For eight different objects of known mass, add each object on the piston and measure the height of the piston for each object.
(ii) Ptot = Patm + (mp + Nm0)g/A = Patm + (mp + Nm0)g/πr2 where N is the number of objects on the piston and r is the radius of the piston.
(iii) ρ = Mg /V Mg /(πr2h)
(iv) According to the ideal gas law, pressure is proportional to 1/V. Because the mass of this gas is constant, pressure is, therefore, directly proportional to density. The graph does not show a linear relationship between density and pressure, so the gas is not ideal.
(b) When the balloon goes deeper in the fluid, the pressure increases. This will cause the volume of the balloon to decrease.
(c) ∑F = 0 = FB – Wballoon – Wgas – Fstudent
FB = ρWVbg
W = ρgVbg + mbg
Fstudent = ρwVbg – ( ρgVbg+mbg)
Question 3
An electromagnet produces a magnetic field that is uniform in a certain region and zero outside that region. The graph above represents the field as a function of the current in the electromagnet, with positive field directed out of the page and negative field directed into the page.
(a) The current in the electromagnet is set at 0.5I1. When a charged particle in the region moves toward the top of the page, the force exerted on it by the field is FB toward the left, as shown above. What changes to the current in the electromagnet could make the magnitude of the force exerted on the particle equal to 2FB and the direction of the force to the right? Support your answer using physics principles.
A circuit is made by connecting an ohmic lightbulb of resistance R and a circular loop of area A made of a wire with negligible resistance. The circuit is placed with the plane of the loop perpendicular to the field of the electromagnet, as shown above on the left. The magnetic field changes as a function of time, as shown in Graph 2. The bulb dissipates energy during the interval t1 < t <t3. Graph 3 below shows the cumulative energy dissipated by the bulb (the total energy dissipated since t = 0) as a function of time.
(b) The original bulb is replaced by a new ohmic lightbulb with a greater resistance, but everything else stays the same. How would the cumulative energy graph for the new bulb be different, if at all, from Graph 3 above? Support your answer using physics principles.
(c) The new lightbulb is removed and replaced by the original lightbulb. The magnetic field now changes from 2B1 to −2B1 during the same interval t1 < t <t3. A new cumulative energy graph is created for this situation. How would the new graph be different, if at all, from Graph 3 ? Support your answer using physics principles.
(d) A student derives the following expression for the cumulative energy dissipated by the original bulb during the interval t1 < t <t3 and with the original change in magnetic field shown in Graph 2.
\(Energy = \frac{A^{2}B_{1}R}{4(t_{3}-t_{1})}\)
Whether or not the equation is correct, does the functional dependence of cumulative energy on the elapsed time ( t3−t1) make physical sense? Support your answer using physics principles.
Answer/Explanation
(a) The current must change direction and double in magnitude. The graph shows that when the current doubles the magnetic field doubles. When the magnetic field doubles, the magnetic force doubles. Reversing the direction of the current will reverse the direction of the magnetic field, and therefore the direction of the force.
(b) The slope of the energy vs. time graph represents power. Because the induced emf is the same, but resistance is higher, power is lower. Therefore, the slope would be smaller.
(c) The induced emf is larger than it was before because the magnetic field changed by a larger amount in the same time period. Power is proportional to the square of the emf, so the power is larger. Power is the slope of energy vs time, so the slope is greater for the new graph.
(d) The, dependence does make sense: it shows that as the time interval increases, the energy dissipated will decrease. This makes sense : the magnetic field would be changing at a rate of almost O, so the change in flax would be almost 0, the induced emf would be almost 0, thereby the energy dissipated would be almost 0
Question 4
Light and matter can be modeled as waves or as particles. Some phenomena can be explained using the wave model, and others can be explained using the particle model.
(a) Calculate the speed, in m/s, of an electron that has a wavelength of 5.0 nm.
(b) The electron is moving with the speed calculated in part (a) when it collides with a positron that is at rest. A positron is a particle identical to an electron except that its charge is positive. The two particles annihilate each other, producing photons. Calculate the total energy of the photons.
(c) A photon approaches an electron at rest, as shown above on the left, and collides elastically with the electron. After the collision, the electron moves toward the top of the page and to the right, as shown above on the right, at a known speed and angle. In a coherent, paragraph-length response, indicate a possible direction for the photon that exists after the collision and its frequency compared to that of the original photon. Describe the application of physics principles that can be used to determine the direction of motion and frequency of the photon that exists after the collision.
Answer/Explanation
(a) \(\lambda = h/p = h/mv\)
\(v = \frac{\left ( 6.63 * 10^{-34} J.s \right )}{(9.11 * 10^{-31 kg})(5.0 * 10^{-9} m)}\)
\(v = 1.5 _10^{5} m/s\)
(b) Etot = 2(9.11 * 10-31 kg)(3 *108 m/s)2 + (1/2)(9.11 *10-31 kg)(1.5 * 105 m/s)2
Etot = 1.6 * 10-31 J
(c) In order to conserve momentum in the vertical direction, the photon must have a component of its momentum toward the bottom of the page. If the horizontal component of the momentum of the electron after the collision is less than the initial momentum of the photon, then the photon must move toward the right after the interaction. In order to conserve energy, the frequency of the photon after the collision is less than what it was before the collision because it gave some of its energy to the electron.