The First Law of Thermodynamics AP Physics 2 FRQ – Exam Style Questions etc.
The First Law of Thermodynamics AP Physics 2 FRQ
Unit 9: Thermodynamics
Weightage : 15–18%
Exam Style Practice Questions,The First Law of Thermodynamics AP Physics 2 FRQ
Question
Students observe that a graphite rod gets hot when there is an electric potential difference \(\Delta V\) applied across it that causes an electric current I in it. The graphite rod is placed in an apparatus that consists of a clear plastic container with a lid, as shown above. The lid is equipped with electrical connectors and an opening that can be sealed around an inserted sensor. The graphite rod is connected to the electrical connectors by wires and sealed inside the container so that all the energy emitted by the rod goes into heating the air in the container. The teacher tells the students that in this situation the change in the internal energy of the air is equal to (5 2)\(Nk_{B}\Delta T\), where N is the number of molecules and T is the temperature, and the air can be treated as an ideal gas.
(a) Derive an expression for the temperature change of the air as a function of time t as a result of the electrical energy dissipated by the rod and delivered to the air in the container. Express your answer in terms of I, \(\Delta V\), N, and physical constants, as appropriate. Assume that the temperature of the graphite rod remains constant while the air is being heated.
The students are asked to design an experiment using the apparatus shown to investigate this heating. The students have an ammeter, a voltmeter, a fixed DC power supply, a stopwatch, an electronic temperature sensor, and a pressure sensor. Assume that the electrical connectors and connecting wires have negligible resistance.
(b) Outline an experimental procedure that can be used to gather data to determine how the air temperature in the container depends on the electrical energy delivered to the rod. Indicate the measurements to be taken and how the measurements will be used to obtain the data needed. On the diagram on the previous page, show how the container will be connected to instruments to take the necessary measurements.
(c) On the axes below, sketch the line or curve you predict will represent a plot of the temperature of the air in the container as a function of electrical energy delivered to the rod.
(d) i. On the axes below, sketch a line or curve you predict will represent a plot of the pressure of the air in the container as a function of time. Clearly label the sketch as \(R_{1}\).
Explain why your graph has this shape.
ii. The rod is now replaced by a second graphite rod that has twice the length but the same radius. The potential difference across the new rod is the same as that across the original rod. On the axes in part (d)(i), sketch a line or curve you predict will represent a plot of the pressure of the air in the container as a function of time for the second rod. Clearly label the sketch as \(R_{2}\). Compare this graph to the graph from part (d)(i) and explain why it is the same or different.
(e) Another group of students performing this experiment notices a gap in the seal of the container opening and thinks that some gas has leaked out of the container. If this is true, how would this group’s graph of air temperature as a function of electrical energy compare to the graph you drew in part (c)?
Answer/Explanation
Ans:
(a) \(\Delta E=I\Delta Vt\)
\(I\Delta Vt=5/2 Nk_{B}\Delta T\)
\(\Delta T=2\Delta VIt/5Nk_{B}\)
(b) For a valid description of the setup and procedure, including a diagram
For measuring current and potential difference, with symbols defined as needed
For measuring the temperature change of the air, with a symbol defined as needed
For a description of how the measurements will be used to calculate the energy
(c) 
(d) i) 
ii) 
(e) The new graph will have a different slope, and with no indication that the shape of the new graph would be different.
Question: (10 points, suggested time 20 minutes)
Two moles of a monatomic ideal gas are enclosed in a cylinder by a movable piston. The gas is taken through the thermodynamic cycle shown in the figure above. The piston has a cross-sectional area of 5 × 10-3 m2 .
(a)
i. Calculate the force that the gas exerts on the piston in state A, and explain how the collisions of the gas atoms with the piston allow the gas to exert a force on the piston.
ii. Calculate the temperature of the gas in state B, and indicate the microscopic property of the gas that is characterized by the temperature.
(b)
i. Predict qualitatively how the internal energy of the gas changes as it is taken from state A to state B. Justify your prediction.
ii. Calculate the energy added to the gas by heating as it is taken from state A to state C along the path ABC.
(c) Determine the change in the total kinetic energy of the gas atoms as the gas is taken directly from state C to state A.
Answer/Explanation
Ans:
\(P = \frac{F}{A}\) FA = PA = 1 × 105 Pa × 5 × 10-3 m2
FA = 500 N
The collision of the gas atoms with the piston allow the gas to exeit a force or the piston because the molecules are in rapid, random motion, and ave frequently colliding with the walls of the cylinder and the piston. When molecules hit the piston in rapid random motion, a force is exerted.
PV = nRT T= \(\frac{PV}{nR}\) = \(\frac{1.0\times 10^{5}Pa \times 0.10m^{3}}{2nd\times 8.31\frac{J}{Molk}}\)
T = 602 K
The average kinetic energy of the molecules of gas is characterized by the temperature, and average kinetic energy relates to the microscopic property of the speed of the gas molecules.
(b)
i.
The internal energy of the gas increases as it is taken from state A to state B because the pressure remains constant and volume increases, meaning that temperature increases. When temp increases, internal energy increases because \(U = \frac{3}{2}nRT.\)
ii.
ΔV = Q + W \(\Delta V = \frac{3}{2}nR\Delta T\) \(\Delta T = \frac{P_{C}V_{C}-P_{A}V_{A}}{nR}\)
1496J = -6000 J + Q ΔV = 1, 496 J ΔV = 60 K
W = P Δ V
W = 1×105Pa×0.06 m3
W = -6000 J
ii.
Δ V = \(\frac{3}{2}nR\Delta T\) Δ T = TC – TA = \(\frac{P_{A}V_{A}}{nR}-\frac{P_{C}V_{C}}{nR}\)
Δ V = \(\frac{3}{2}.2nd 8.31\frac{J}{molk}. 60K\)
Δ T = \(\frac{0.5\times 10^{5}Pa\times 0.10m^{3}}{2nd \times 8.31\frac{J}{molk}}-\frac{1.0\times 10^{5}Pa\times 0.04m^{3}}{2nd \times 8.31\frac{J}{molk}}\)
Δ V = 1496 J Δ T = -60 K