PHYSICS 313
Second midterm November 6 2002.

Answer all three questions. Allowed aids 1 double-sided formula sheet. Calculator. Some potentially useful formulas are given on the second page of the exam.
Problem 1:
A container has initially 2 compartments each with volume 1 liter. One compartment contains 1 mol and the other 2 mol of the same monatomic ideal gas as the first compartment. The temperature is initially 300K. The wall separating the two compartments is removed.
a: What is the final pressure and temperature?
b: What is the change in entropy when the gases in the two containers are mixed?
c:Would you have got a different result in b: if the gas had been diatomic?
Problem 2:
A refrigerator pumps heat from the inside compartment into the room in which it is located. Let $T_C(T_H)$ be the inside (outside) temperature, $Q_C$ the heat remove and $W$ the work needed to do this. The coefficient of performance is $COP=Q_C/W$.
a: Use the second law of thermodynamics to derive an upper limit to the achievable $COP$.
b: An actual fridge operates at 1/2 the ideal performance. It consumes constant power $J$ and heat seeps into the cold compartment at the rate $\alpha(T_H-T_C)$, with $\alpha$ a constant. What temperature will be reached inside the fridge if the temperature $T_H$ is constant?
Problem 3: The entropy of liquid water at 298 K atmospheric pressure per mol is S=69.91 JK$^{-1}$.
a: What is the change in entropy per mol if water is heated from 298 K to 340 K at constant pressure.
b: Estimate the change in the Gibbs free energy when 1 mol water is heated from 298 K to 340 K at constant pressure.
Some formulas:
In the formulas $P_iV_i^\gamma=P_fV_f^\gamma,\; U=\frac{f}{s}nRT$, we have $\gamma=5/3, f=3$ for a monatomic gas, while $f=5,\;\gamma=1.4$ for air.

$$G=\mu N=U+PV-TS;\;\; H=U+PV$$

$$dG=-SdT+VdP+\mu dN$$

$$dH=TdS+VdP+\mu dN$$

Sackur tetrode formula for monatomic ideal gas

$$S=Nk_B[\ln(\frac{V}{v_qN})+\frac{5}{2}]$$

$$v_q=\left(\frac{h^2}{2\pi mkT}\right)^{3/2}$$

$R=8.315\; {\rm J mole}^{-1}{\rm K}^{-1}$ in ideal gas law $PV=nRT$.
$N_A=6.022\times10^{23},\; k_B=1.381\times 10^{-23}{\rm J\; K^{-1}}.$
1 atm$= 1.013$ bar $=1.013\times 10^5$ N m$^{-2}$.
$C_P=4.18$ kJ kg$^{-1}$ for water.