Students are sometimes unclear about which of the equations taught in the topic of Thermal Physics apply to ideal gases and which apply to all systems (whether ideal or real gas, even liquids and solid). The following table should help to clarify:
Applies to Ideal Gas only
Applies to all systems
Gas Laws
$$pV=nRT$$
$$pV\approx nRT$$ only for gases at low pressure and high temperatures
Average Kinetic Energy
$$=\dfrac{3}{2}kT$$
$$\propto T$$
Internal Energy
$$U$$ = sum of KE of molecules $$U=\dfrac{3}{2}NkT=\dfrac{3}{2}nRT=\dfrac{3}{2}pV$$
After going through the topic of Dynamics in the A-level syllabus recently, it strikes me that there are several common mistakes or misconceptions that students whom I have taught over the years have always struggled with. I shall lay them out here in the hope that students who read this do not repeat the same mistakes.
1. Mixing up free-body diagrams involving multiple bodies in motion
Students are often confused about the forces in drawing free-body diagrams, especially so when they have to consider the different parts of multiple bodies in motion. A common mistake that students make is to draw all the forces, including the internal forces within a system, on a single diagram.
To solve a two-body dynamics problems, students need to learn how to draw the 3 possible free-body diagrams. A detailed set of instructions is found in an earlier post.
2. Adding instead of subtracting momenta to find impulse
Students who do not have a strong foundation in Math may not appreciate the fact that the addition and subtracting of vectors is not done by simply adding or subtracting the magnitudes. Sometimes, students are asking to find the force acting on an object when given the initial and final momenta. The correct way to find the vector of the change in momentum is to place the tails of the initial and final momenta together, and to join the arrowhead of the initial momentum with that of the final momentum with a third vector in that direction.
Vector triangle to find change in momentum
They are often confused between finding the change in momentum and the act of adding two vectors together. When that happens, what teachers see is the following triangle instead.
Wrong vector diagram to find change in momentum
While the magnitude of the vector found in both diagrams are the same, students will get the wrong answer if asked for the direction of the change in momentum. This is actually an important notion because the direction of change in momentum is really the direction of the net force applied on the body.
3. Using wrong signs for momentum for head-on collisions
In A-level physics, in the study of collision problems, we will deal only with head-on collisions where two objects collide along the same line. Even then, students can get confused with the sign convention associated with the direction of motion of the objects.
Applying Conservation of Momentum
For a two body head-on collision problem, always start by writing down the following equation for the conservation of momentum:
[latex]m_Au_A +m_Bu_B = m_Av_A + m_Bv_B[/latex]
where [latex]m_X[/latex] is the mass of X, [latex]u_X[/latex] is the initial velocity of X, [latex]v_X[/latex] is the final velocity of X, where X is either A or B.
Follow this up by deciding on which direction you want to define as positive. For example, if you decide that the rightward direction is positive, the value of initial velocity for mass A is positive and that for mass B is negative.
What students sometimes do is to include the negative signs into the equation before substitution, e.g. [latex]m_Au_A +m_B(-u_B) = m_Av_A + m_Bv_B[/latex]. This causes confusion as they do not know where the directions of the final velocities are and hence, do not know which vectors to assign as negative.
What we should do instead is to determine the final direction of the velocities after we have found the values. Before we are able to do so, we shall assume that the final velocities are to the right. If the final value of velocity of a mass is negative, it means that its final velocity is to the left.
Substitute the values into equation.
Assuming that the initial speeds are both 4.0 m s-1;, the mass of A is 3.0 kg, mass of B is 5.0 kg, and that the collision is perfectly inelastic, we have:
Here are some interesting lecture demonstrations on adiabatic thermodynamic processes you can carry out. In an adiabatic process, there is no heat transfer between the system and other systems (including its environment.) According to the First Law of Thermodynamics ($$\Delta U=Q+W$$), where Q = 0, a compression of a gas which is associated with work being done on the gas will cause the internal energy and hence, the temperature of the gas to rise. On the other hand, when an expansion of a gas takes place, the gas will cool down.
This is a game that I play with my classes in the beginning of the course after they have attended the first lecture on base units. It serves as an ice-breaker since this is usually when I first meet them, as well as a recap of what they have learnt.
The rules are stated in the slideshow.
1. Each student is assigned a base unit out of 3 possible ones, kg, m, s. Note that the other 4 base units, A, mol, K and cd are not given because they will be covered in later topics.
2. For a class of 25, the following distribution is recommended:
a. kg – 5
b. m – 10
c. s – 10
3. Students will be asked to form groups when shown a Physical Quantity. Those with a negative power, e.g. the two “seconds” in kg m s–2, will have to sit/squat down.
4. Students who did not form a group will be ousted and the last few left in the game will win prizes.
