Subfield: Thermodynamics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Extra Supplementary Reading Materials:

A vacuum system consists of a chamber of volume $V$ connected to a vacuum pump that is a cylinder with a piston that moves left and right. The minimum volume in the pump cylinder is $V_{0}$, and the maximum volume in the cylinder is $V_{0}+\Delta V$. You should assume that $\Delta V \ll V$.



<img_4356>



The cylinder has two valves. The inlet valve opens when the pressure inside the cylinder is lower than the pressure in the chamber, but closes when the piston moves to the right. The outlet valve opens when the pressure inside the cylinder is greater than atmospheric pressure $P_{a}$, and closes when the piston moves to the left. A motor drives the piston to move back and forth. The piston moves at such a rate that heat is not conducted in or out of the gas contained in the cylinder during the pumping cycle. One complete cycle takes a time $\Delta t$. You should assume that $\Delta t$ is a very small quantity, but $\Delta V / \Delta t=R$ is finite. The gas in the chamber is ideal monatomic and remains at a fixed temperature of $T_{a}$.



Start with assumption that $V_{0}=0$ and there are no leaks in the system.

Question: a. At $t=0$ the pressure inside the chamber is $P_{a}$. Find an equation for the pressure at a later time $t$.

Solution:
During each cycle, the system sucks gas out of the chamber and pushes it into the atmosphere. Since $V_{0}=0$, the inlet valve opens the moment the piston starts moving to the left. When the piston is all the way to the left, a fraction $\Delta V /(V+\Delta V)$ of the gas is in the cylinder. As the piston moves to the right, all of this gas is pushed out, so after a single cycle,



$$

P_{f}=P_{i}\left(\frac{V}{V+\Delta V}\right)

$$



and in general,



$$

P(t)=P_{a}\left(\frac{V}{V+\Delta V}\right)^{t / \Delta t}

$$



While this is technically correct, it can be simplified significantly. Write



$$

P(t)=P_{a}\left(1+\frac{\Delta V}{V}\right)^{-t / \Delta t}=P_{a}\left((1+x)^{1 / x}\right)^{-R t / V}

$$



where $x=\Delta V / V \ll 1$. Then using the definition of $e$,



$$

e=\lim _{x \rightarrow 0}(1+x)^{1 / x}

$$



we have



$$

P(t)=P_{a} e^{-R t / V}

$$

Final answer: $P(t)=P_{a} e^{-R t / V}$