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Chapter 12: Problem 5

The Boltzmann entropy equation is (a) \(S=\frac{K}{\ln W}\) (b) \(S=K^{r} \ln \frac{P_{2}}{P_{\mathrm{J}}}\) (c) \(S=K^{\prime} \ln \frac{V_{\mathrm{J}}}{V_{2}}\) (d) \(S=K^{\prime \prime} \ln W\)

Short Answer

Expert verified
The correct Boltzmann entropy equation is \(S = K^{\prime \prime} \ln W\), which corresponds to answer choice (d).

Step by step solution

01

Review the Boltzmann Entropy Equation

Boltzmann entropy equation is classified from the Boltzmann principle, which relates the entropy \(S\) of an ideal gas to the quantity \(W\), the number of microstates corresponding to a given macrostate, with \(K\) being the Boltzmann constant: \(S = K*\ln W\). Therefore, looking at the choices given:
02

Analyze Each Given Equation

(a) \(S=\frac{K}{\ln W}\) does not match the correct formula, because in the right equation \(K\) is multiplied with \(lnW\), not divided. (b) \(S=K^{r} \ln \frac{P_{2}}{P_{\mathrm{J}}}\) includes pressure \(P\), which is not part of the original Boltzmann entropy equation. (c) \(S=K^{\prime} \ln \frac{V_{\mathrm{J}}}{V_{2}}\) This equation includes volume \(V\), which is not part of the original Boltzmann entropy equation. (d) \(S=K^{\prime \prime} \ln W\) This equation matches the structure of the Boltzmann entropy equation. The notation \(K^{\prime \prime}\) is presumably just another name for the Boltzmann constant. However, if it represents a different value, this wouldn't be correct.

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Most popular questions from this chapter

The third law of thermodynamics may be stated as follows: (a) At absolute zero the entropy of all perfectly crystalline solids tends to decrease. (b) The entropy of a perfectly crystalline solid may be taken as zero at absolute zero. (c) The entropy of every substance is zero at \(0 \mathrm{~K}\). (d) The entropy of a substance is related to its heat capacity.

From the third law, the entropy of a substance at temperature \(T\) is given as, (a) \(S_{T}=\frac{1}{3} C_{P}\left(\right.\) at \(\left.T_{1}\right)+\int_{T_{1}}^{T} \frac{C_{p}}{T} d T\) (b) \(S=C_{p}\left(\right.\) at \(\left.T_{1}\right)+\int_{\tau_{1}}^{T} C_{p} \ln d T\) (c) \(S=C_{p} \ln T+\int_{\tau_{1}}^{T} \frac{C_{p}}{T} d T\) (d) none of the above

Residual molar entropy is given by the expression (a) \(S=R \ln W\) (b) \(S=N \ln W\) (c) \(S=\frac{R}{N} \ln W\) (d) \(S=n R \ln W\)

The Nernst heat theorem can be mathematically stated as (a) \(\lim _{T \rightarrow 0} \frac{d(\Delta G)}{d T}=\lim _{T \rightarrow 0} \frac{d(\Delta H)}{d T}=0\) (b) \(\lim _{T \rightarrow 0} \frac{d(\Delta H)}{d T}=\lim _{T \rightarrow 0} \frac{d(\Delta S)}{d T}=0\) (c) \(\lim _{T \rightarrow 0} \frac{d(\Delta V)}{d T}=\lim _{T \rightarrow 0} \frac{d(\Delta A)}{d T}=0\) (d) \(\lim _{r \rightarrow 0} \frac{d(\Delta A)}{d T}-\lim _{r \rightarrow 0} \frac{d(\Delta H)}{d T}\)
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