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PHYSICS 313 Midterm November 5 2003.
Answer all questions. Allowed aids 1 double-sided formula sheet. Calculator.
Problem 1:
The entropy of one mol of argon at 298 K is 154.84 J/K.
a: Assume that argon is a monatomic ideal gas. Use the information provided to find the quantum volume at 298 K in the Sackur Tetrode equation

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

b: What is the entropy of 1 mol of Argon at 400 K, if the pressure remains 1 bar? You may assume that the quantum volume $v_q$ is proportional to $T^{-3/2}$.
Problem 2:
A container has initially 2 compartments each with volume 1 liter. Each compartment contains 1 mol of the same monatomic ideal gas. The temperature of the two compartments are initially 300K and 400 K. 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 3:
a: The pressure of 1 mol of liquid water is increased isothermally from 1 bar to 300 bar at T=298 K. Estimate the change in the Gibbs free energy neglecting the change in volume.
b: The pressure 1.5 mol of an ideal gas is increased isothermally from 1 bar to 300 bar. Estimate the change in the Gibbs free energy if the temperature is 298 K. (now you cannot neglect the change in volume).
c: In the electrolysis of water

\begin{displaymath}H_2O ->H_2 +\frac{1}{2}O_2\end{displaymath}

2 electrons pass through the circuit per formula unit. Will the minimum voltage required increase or decrease if the electrolysis takes place under high pressure? Justify your answer. Some formulas
$R=8.315$ J mol$^{-1}$ K$^{-1};\;\;$ $N_A=6.022\times 10^{23};\;\; k_B=1.381\times 10^{-23}$ J K$^{-1}.$
1 atm $=1.013$ bar $=1.013\times 10^5$ N m$^{-2}$
Ideal gas: $PV=nRT;\;\; U=\frac{f}{2}nRT,;\;\;C_P=C_V+nR$
$f=3$ for monatomic gas; $f\approx 5$ for diatomic gas
The density of liquid water is $10^3$ kg m$^{-3}$. The molecular weight of water is 18 g mol$^{-1}$.

\begin{displaymath}G=N\mu=U-TS+PV\end{displaymath}


\begin{displaymath}dG=-SdT+VdP+\mu dN; \;\; d\mu=-sdT+vdP\end{displaymath}




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Birger Bergersen 2003-11-12