# Tag Info

3

That $C$ is the specific heat for the given cycle, i.e. $$dQ=nCdT$$ This is for $n$ moles of gas.(not the $n$ you stated in question) I will assume $$PV^z=\text{constant}$$ $$nCdT=dU+PdV$$ $$\int nCdT=\int nC_vdT+\int PdV$$ We will integrate it using Pranjal's method : $$nC\Delta T=nC_v \Delta T+\int \frac{PV^z}{V^z}dV$$ As numerator is a constant, ...

3

Without a doubt, it is the zeroth law of thermodynamics, as it defines an equivalence relation. It states that If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.

2

John Rennie's answer is correct for a DC series connected motor and, almost certainly, this is the kind of motor you (the OP) are talking about. An interesting way of writing John's answer "backwards" is that you have just observed the reason why the most powerful traction motors are exactly this kind of motor - almost all DC train and tram motors are ...

2

The simplest reaction deuterium and tritium. Tritium is common in big labs (like NIF, JET, Omega) [1]. Tritium sucks - practically speaking. It is expensive, radioactive, and hard to stockpile. Omega spent millions and years on a tritium facility. It may even never be used in fusion power [2]. The next easiest reaction is deuterium with itself. This ...

2

By definition a reversible adiabatic system has $dQ = 0$. We also know the following from the Clausius Theorem : $dS = \frac{dQ}{T}$ Then it is easy to see that there can be no change in entropy. Note that irreversible adiabatic systems CAN see a change in entropy because in that case the above equation is no longer an equality but an inequality : $dS ... 2 The simple answer: Satellites do feel this force, but obviously don't get ripped apart. The tidal forces are simply too small (for the satellites' materials) to actually rip them apart. The Why: Tidal forces happen because one side of an object feels such a larger huge difference in force than the other side. The magnitude of the force not only has to deal ... 2 I believe the author is thinking here of things like electrochemical work that don't involve a change of volume (but rather, in this case, moving a charge through a potential difference). In this case a full thermodynamic description involves the chemical potential$\mu$, as a later equation shows. (I agree that the description of “different free energies ... 2 I don't think I quite understand your question, but I'll do my best. In Thermodynamics, pressure is defined in a bevy of ways. If we look at the Thermodynamic Identity: $$dU = TdS - PdV + \mu dN$$ (where$U$is the Energy,$T$is the Temperature,$S$is the Entropy,$P$is the Pressure,$V$is the Volume,$\mu$is the Chemical Potential, and$N\$ is the ...

1

If I understand the question, you are wondering how to justify the statement that a (reverible) adiabatic process is isentropic from the point of view of statistical mechanics (the classical thermodynamics definition makes sense to you). Let us then start with the entropic fundamental relationship, S = S (U, V, N), where U stands for energy, V for volume, N ...

1

If you rearrange the ideal gas law to be expressed in terms of pressure $$P=\frac{NRT}{V}\qquad\Rightarrow\qquad \frac{T_1}{V_1}=\frac{T_2}{V_2}$$ where the right hand equation assumes it is an isobaric process with no mass exchange. So, in an isobaric process temperature and volume vary inversely. If the volume decreases then the temperature must go up. ...

Only top voted, non community-wiki answers of a minimum length are eligible