# Tag Info

7

The formula for momentum is not $p=mv$ but it is $\vec p=m\vec v$. This being said, after an explosion, the velocities of the fragments have increased and so is the kinetic energy of the entire system (chemical energy $\rightarrow$ kinetic energy in the explosion) but the net momentum does not change. If the system had zero net momentum before the explosion, ...

4

How is it possible to detect a single photon without making any change to it? In general, if you have detected a photon in your experimental apparatus, you have changed it drastically. It may have disappeared completely, as in this bubble chamber picture: The colored diagram shows the photon in the picture that has materialized into an electron ...

3

This might be easier to understand if you think of a simpler case where the "firecracker" is only made up of two masses. Let's say the two masses are 40 g each, and when the device explodes it sends each mass going the opposite way at 130 m/s. Before the explosion, the combined 80 g mass is moving to the right at 1 m/s. Its momentum is therefore 80 g m / ...

3

Excludes the system as a whole means you disregard the kinetic energy associated with the translation or rotation of the body as a whole. For instance that associated with the center of mass motion. In case of internal energy you focus on the interior, like motion due to the atoms and molecules of which the object is made up of.

2

A system is a set of interacting or interdependent components that act as a whole, and the system's boundary delineates the system from its surroundings or environment in which it exists. Wikipedia offers a good definition here. If I'm understanding your question, you are asking whether objects, defined by their energy states, can be considered a system. ...

2

Assuming that $V\propto n$, since $P$ and $T$ are intensive quantities, the first equation implies that $a$ it's proportional to $n^{\frac{1}{3}}$. As you can see this also works good with the other two. I'm not sure that is always the case that $V\propto n$, though. Since $n$ is directly proportional to the mass, which is directly proportional to the ...

2

The average energy is $$\overline{U}=-\frac{\partial}{\partial\beta}\log(Z)=-\frac{1}{Z}\frac{\partial Z}{\partial\beta}=\frac{\sum_pg_p\epsilon_p\exp(-\epsilon_p\beta)}{\sum_pg_p\exp(-\epsilon_p\beta)}=\sum_p\epsilon_pP_p$$ where $P_p$ is the probability of being in the $p^\text{th}$ state. Multiplying this by $N$ (the total number of particles) and noting ...

2

The change in electronic excitation represents both a potential and a kinetic energy term in classical physics, but there is no simple correspondence to classical physics terms, when you are looking at quantum systems. All we really care about is the total energy difference between electronic states. Those energy differences correspond to the energies at ...

2

What constituent of internal energy does an electron excitation represent? You can think of electrons as just like planets orbiting the sun and get the correct answer to this question. An electron in a higher energy level has less kinetic energy, but more potential energy as it is (generally) farther from the nucleus. The net result is more energy. ...

2

Who is right depends on what your model for the gas is. In short, you are right for models that incorporate inter-particle interactions. The internal energy of an ideal gas, which is a model that neglects interactions between particles, can be written only in terms of the number of particles $N$ in the gas sample and the absolute temperature $T$ of the ...

2

I don't blame you for being confused by this because the notation in which the first law of thermodynamics is often written ($\mathrm{d}U = \mathrm{d}Q - \mathrm{d}W$) is actually rather inconsistent. In your notation, the specific heat would be $$C_V = \biggl(\frac{\omega_q}{\mathrm{d}T}\biggr)_V$$ It's just conventional to write $\omega_q$ as ...

2

And since K increased v increased and thus p=mv must increase? For an individual particle, yes. For a system of particles, no. Consider two identical particles, co-located and initially at rest. At initial rest, the total kinetic energy is zero and the total momentum is zero. Now, due to some mechanism, the particles are sent in opposite directions ...

2

Great question; I myself got confused for a moment there. I'm gonna try to be somewhat thorough, so bear with me. First consider the differential form of the first law of thermodynamics which holds for any quasi-static process. $$dE = \delta Q - \delta W$$ For an adiabatic process, $\delta Q = 0$ by definition, so one obtains $$dE = -\delta W$$ On ...

2

The act of measurement causes a quantum system to collapse into an eigenstate of the operator associated with the measurement. So unless the photon wave function is in an eigenstate of the detection operator, it will be changed. Unless a system is in an eigenstate of the hamiltonian it will not have a definite energy. If one measures the energy of the ...

1

The internal energy of an ideal gas can only change if heat is added or removed from the system, or if the system does some work. Neither is the case in this example so the change in internal energy is zero. The final pressure is, as you say, just half the initial pressure. I'm not sure why they give you the specific heats, but you can use them to work out ...

1

When you lift a box of an ideal gas you are not doing any work on the gas, so its internal energy remains constant. However you are increasing the gravitational potential energy of the box and the gas. That's where the energy you put goes.

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There is a difference between an electron gaining energy and the whole system (atom) gaining energy. Electron orbital energy levels are quantized and make discreet jumps. For this reason "centripetal force" for electrons doesn't make much sense. There is an electron orbital binding energy though which is equivalent to the "work function" in your question. ...

1

I think you have a misunderstood, the quantity entalphy is a definition, it comes from the second law (conserved energy, if there is constant pressure ($\delta W =pdV$) $dU=\delta Q-\delta W \Rightarrow dU=\delta Q-pdV$ from the chain rule $d(pV)=Vdp+pdV \Rightarrow pdV=d(pV)-Vdp$, so \$ dU=\delta Q -d(pV)+Vdp \Rightarrow d(U+pV)=\delta Q+Vdp ...

1

Rubber consists of many long-chain polymers. In an unstressed sample, these are randomly arranged. As a mental model think of them as anchor chains, where the angle between each link is entirely random - the overall polymer is in essence a random walk through the medium. Now, you pull on it. The net result is to better align the backbone of the polymer ...

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