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Hot answers tagged internal-energy

6

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, ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 / ...

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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 ...

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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 ...

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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 ...

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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. ...

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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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