There is an intuitive reason why a vibration is IR-active only if it involves a change in the dipole moment. Recall that the typical wavelength of IR radiation ($\sim 10\mu\mathrm{m}$) is much larger ...

Usually this part of thermodynamics is not presented in the most efficient way. To derive the expression for, e.g., $C_P - C_V$, one goes through a series of steps involving Maxwell's relations and ...

The relation $dG = VdP - SdT$ implies that the Gibbs free energy of the system depends only on the two variables $T$ and $P$, e.g., as in the case of a single-component gas consisting of a fixed ...

As far as the uncertainty relation between $L_{z} = -i\hbar \partial/\partial\phi$ and $\phi$ is concerned, there is a simple way to understand why we cannot use the commutation relation $[\partial/\... View answer Accepted answer 5 votes Whatever one observes (whether it is a liquid, a gas, or a mixture thereof) is the state of maximum entropy for the particular values of the internal energy and volume the system currently has. As OP ... View answer 4 votes What OP cited is the derivation of the Boltzmann distribution starting from the micro-canonical ensemble, characterized by a fixed energy and a fixed number of constituent particles. In mathematical ... View answer 2 votes Your$\Delta G$is really$\Delta G_{\mathrm{reaction}}$, and the reaction does not happen alone: it is accompanied by charging of electrodes. The Gibbs free energy of charging is just the associated ... View answer 2 votes As Boltzmann taught us, (equilibrium) thermodynamics1 is a natural consequence of the statistical behavior arising in any macroscopic system, regardless of the details of the microscopic laws of ... View answer 2 votes The following is a solution alternative to OP's own. As OP correctly observed, any$C_3$rotation belonging to the group$T_d$is similar to the$C_3$rotation about$\hat{\textbf{n}} = (\hat{\textbf{...

The entropy function is of the form $S(T,p,n) = n\, s(T,p)$, where $s(T,P)$ is the entropy per mole. The surrounding is always much, much larger than the system. Hence, for the surrounding, $n$ is an ...

It is often said that $U = 5nRT/2$ for a diatomic gas (c.f. $U = 3nRT/2$ for a monatomic gas). According to this formula, the internal energy should not change in the reaction you considered, provided ...
The fact that $Q_{\mathrm{initial}} = Q_{A}Q_{B}$ implies that the two subsystems, respectively consisting of gas A and gas B, are independent. Because \$\log Q_{\mathrm{initial}} = \log Q_{A} + \log ...