GATE 2024 PH – Question 59
An ideal gas has $Q=\frac1{N!}[V/\lambda(T)^3]^N$, where $\lambda$ is thermal de Broglie wavelength. Use $\ln N!=N\ln N-N$. For density $N/V=2.5\times10^{25}$ m−3, $e^{\mu/(k_BT)}\lambda(T)^{-3}\times10^{-25}$ is _____ m−3 (one decimal place).
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 2.49 to 2.51
Explanation
The grand-canonical quantities follow from the canonical partition function through the chemical potential:
$$\mu=-k_BT\left(\frac{\partial\ln Q}{\partial N}\right)_{V,T}.$$
**Take the logarithm** of $Q=\dfrac1{N!}\left[\dfrac{V}{\lambda^3}\right]^N$ using Stirling's approximation $\ln N!\approx N\ln N-N$:
$$\ln Q=N\ln\frac V{\lambda^3}-N\ln N+N=N\ln\frac{V}{N\lambda^3}+N .$$
**Differentiate** with respect to $N$:
$$\frac{\partial\ln Q}{\partial N}=\ln\frac{V}{N\lambda^3}+1-1=\ln\frac{V}{N\lambda^3}.$$
$$\mu=-k_BT\ln\frac{V}{N\lambda^3}=k_BT\ln\left(\frac N V\lambda^3\right).$$
**The requested combination:**
$$e^{\mu/k_BT}\lambda^{-3}=\frac NV=2.5\times10^{25}\text{ m}^{-3}.$$
So the value of $e^{\mu/k_BT}\lambda^{-3}\times10^{-25}$ is **2.5**.