Problem set 75

1. The internal energy of $n$ moles of a gas is given by $E = \frac{3}{2}nRT- \frac{a}{V}$, where $V$ is the volume of the gas at temperature $T$ and $a$ is a positive constant. One mole of the gas in state $(T_1, V_1)$ is allowed to expand adiabatically into vacuum to a final state $(T_2, V_2)$. The temperature $T_2$ is
1. $T_1+Ra\left(\frac{1}{V_2}+\frac{1}{V_1}\right)$
2. $T_1-\frac{2}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$
3. $T_1+\frac{2}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$
4. $T_1-\frac{1}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$

According to first of thermodynamics $$dU=dQ+dW$$ For adiabatic process there is no heat transfer. Hence, $dQ=0$ $$dU=dW$$ For a free expansion of gas in vacuum, work done is zero, $dW=0$ $$dU=E_2-E_1=0$$ $$E_2=E_1$$ $$\frac{3}{2}nRT_2- \frac{a}{V_2}=\frac{3}{2}nRT_1- \frac{a}{V_1}$$ $$\frac{3}{2}nRT_2 =\frac{3}{2}nRT_1- \frac{a}{V_1}+\frac{a}{V_2}$$ $$T_2 =T_1+\frac{2a}{3nR}\left( \frac{1}{V_2}-\frac{1}{V_1}\right)$$ for one mole $n=1$ $$T_2 =T_1+\frac{2a}{3R}\left( \frac{1}{V_2}-\frac{1}{V_1}\right)$$

Hence, answer is (C)

2. A monatomic crystalline solid comprises of $N$ atoms, out of which $n$ atoms are in interstitial positions. If the available interstitial sites are $N'$, the number of possible microstates is
1. $\frac{(N'+n)!}{n!N!}$
2. $\frac{N!}{n!(N+n)!}\frac{N'!}{n!(N'+n)!}$
3. $\frac{N!}{n!(N'-n)!}$
4. $\frac{N!}{n!(N-n)!}\frac{N'!}{n!(N'-n)!}$

$\frac{N!}{n!(N-n)!}\frac{N'!}{n!(N'-n)!}$

Hence, answer is (D)

3. A system of $N$ localized, non-interacting spin $1/2$ ions of magnetic moment $\mu$ each is kept in an external magnetic field $H$. If the system is in equilibrium at temperature $T$, the Helmholtz free energy of the system is
1. $Nk_BT\ln{\left(\cosh{\frac{\mu H}{k_BT}}\right)}$
2. $-Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)}$
3. $Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)}$
4. $-Nk_BT\ln{\left(2\sinh{\frac{\mu H}{k_BT}}\right)}$

A spin $\frac{1}{2}$ has two possible orientations. (These are the two possible values of the projection of the spin on the z axis: $m_s=\pm1/2$.) Associated with each spin is a magnetic moment which has the two possible values $\pm\mu$. Hence, in a magnetic field the two states are of different energy ($E=-\vec\mu\cdot\vec H$). Hence, single particle partition function is given by \begin{align*} z&=\sum e^{-E/k_BT}\\ &=e^{-\mu H/k_BT}+e^{\mu H/k_BT}\\ &=2\cosh{\frac{\mu H}{k_BT}} \end{align*} Hence, N particle partition function is $$Z=z^N=\left(2\cosh{\frac{\mu H}{k_BT}}\right)^N$$ Helmholtz free energy is given by \begin{align*} F&=-k_BT\ln{Z}\\ &=-Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)} \end{align*}

Hence, answer is (B)

4. The phase diagram of a free particle of mass $m$ and kinetic energy $E$, moving in a one-dimensional box with perfectly elastic walls at $x = 0$ and $x = L$, is given by


For free particle $E=\frac{p_x^2}{2m}$ $$p_x=\pm\sqrt{2mE}$$ Particle bounce forth and back between $x=0$ and $x=L$

Hence, answer is (A)

5. In the microwave spectrum of identical rigid diatomic molecules, the separation between the spectral lines is recorded to be $0. 7143\: cm^{-1}$. The moment of inertia of the molecule, in $kg\: m^2$, is
1. $2.3 \times 10^{-36}$
2. $2.3 \times 10^{-40}$
3. $7.8 \times 10^{-42}$
4. $7.8 \times 10^{-46}$

In rotational spectra separation between two spectral line is $2B=0. 7143\: cm^{-1}$ $$B=\frac{h}{8\pi^2cI}$$ When $B$ is in $cm^{-1}$, $c$ should be taken in $cm\:s^{-1}$. \begin{align*} I&=\frac{h}{4\pi^2c2B}\\ &=\frac{6.62607\times10^{-34}}{4\times3.14^2\times3\times10^{10}\times0.7143}\\ &=7.832\times10^{-46} \end{align*}

Hence, answer is (D)