PHYSICS 313
Second midterm November 5 2001.

Answer all three questions. Allowed aids 1 double-sided formula sheet. Calculator. Some potentially useful formulas are given at the end of the exam.

Problem 1:
A simplified $PV$ diagram for the Joule ideal gas cycle is shown on the figure on the next page. Assume that the working substance is air. The gas is first compressed adiabatically from $P_1$ to $P_2$. It is next heated at constant pressure $P_2$ so that the temperature changes from $T_2$ to $T_3$. The gas then expands adiabatically until the pressure is reduced to the initial pressure $P_1$ and the temperature is $T_4$. Finally the gas is cooled under constant pressure until it returns to its initial state.
a: Find an expression for the thermodynamic efficiency $e=W/Q_{Hot}$ for a heat engine operating according to the Joule cycle in terms of the temperatures $T_1,T_2,T_3,T_4$.
b: Unfortunately, the temperatures are not known, only the pressures. Express the thermodynamic efficiency of the cycle in terms of $P_1$ and $P_2$.
c: Find a numerical answer for the thermodynamic efficiency if $P_1=1$ bar, $P_2=8$ bar.

Problem 2:
One mol of the monatomic ideal gas argon is heated from 296 K to 300 K at constant pressure. The entropy of one mol of argon at 298 K is 154.84 J/K.

a: What is the change in energy?
b: What is the change in entropy?
c: What is the change in enthalpy?
d: What is the change in Gibbs free energy?

Problem 3:
$M_1=10$ kg of water at 90 $^0$C is mixed with $M_2=6$ kg of water at 10 $^0$C.

a: What is the final temperature?
b: What is the change in entropy of the water?

Figure: PV diagram for the Joule cycle in problem 1

Some formulas:
In the formulas $P_iV_i^\gamma=P_fV_f^\gamma,\;U=\frac{f}{2}nRT$, we have $\gamma=5/3,\; f=3$ for a monatomic ideal gas, while $f=5$, $\gamma=1.4$ for air.

$$G=H-TS;\;\;H=U+PV; dG=-SdT+VdP+\mu dN;dU=TdS-PdV+\mu dN$$

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