\( \begin{array}{l}\overline{\Omega_{1}}=\frac{\overline{k_{B} T}}{} \\ \frac{\Omega_{2}}{\Omega_{1}}=l_{\mathrm{n}} \frac{\mathrm{P} \Delta \mathrm{V}}{\mathrm{k}_{\mathrm{B}} \mathrm{T}} \\ \frac{\Omega_{2}}{\Omega_{1}}=\exp \frac{\mathrm{P} \Delta \mathrm{V}}{\mathrm{k}_{\mathrm{p}} T} \\\end{array} \)
Added by Ayushi
Close
Step 1
- \( \Omega_{1} \) and \( \Omega_{2} \) are the initial and final states of a system, respectively. - \( k_{B} \) is the Boltzmann constant, which is a physical constant that relates energy at the individual particle level with temperature observed at the bulk Show more…
Show all steps
Your feedback will help us improve your experience
Kirsty Gledhill and 65 other Intro Stats / AP Statistics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
From the equation $l=a \sin \omega t$ $$ \begin{aligned} &\frac{d l}{d t}=v=a \omega \cos \omega t \\ &\text { So, } \quad w_{t}=\frac{d \nu}{d t}=-a \omega^{2} \sin \omega t, \text { and } \\ &w_{n}=\frac{\nu^{2}}{R}=\frac{a^{2} \omega^{2} \cos ^{2} \omega t}{R} \end{aligned} $$ (a) At the point $l=0, \sin \omega t=0$ and $\cos \omega t=\pm 1$ so, $\omega t=0, \pi$ etc. Hence $$ w=w_{n}=\frac{a^{2} \omega^{2}}{R} $$ Similarly at $l=\pm a, \sin \omega t=\pm 1$ and $\cos \omega t=0$, so, $w_{n}=0$ Hence $w=\left|w_{t}\right|=a \omega^{2}$
Physical Fundamentals Of Mdchanics
Kinematics
$$ \begin{aligned} &Q_{1}=\frac{C_{p}}{R} p_{1}\left(V_{2}-V_{1}\right), Q_{2}^{\prime}=\frac{C_{p}}{R} p_{2}\left(V_{3}-V_{4}\right) \\ &\text { So } \quad \eta=1-\frac{p_{2}\left(V_{3}-V_{4}\right)}{p_{1}\left(V_{2}-V_{1}\right)} \\ &\text { Now } p_{1}=n p_{2}, p_{1} V_{2}^{\gamma} \text { or } V_{3}=n \frac{1}{\gamma} V_{2} \\ &p_{2} V_{4}^{\gamma}=p_{1} V_{1}^{\gamma} \text { or } V_{4}=n^{\frac{1}{\gamma}} V_{1} \\ &\text { so } \eta=1-\frac{1}{n} \cdot n^{\frac{1}{y}}=1-n^{\frac{1}{\gamma}-1} \end{aligned} $$
Thermodynamics And Molecular Physics
The Second Law of Thermodynamics. Entropy
$$ \begin{aligned} &\Delta S=-\frac{m q_{1}}{T_{2}}-m c \ln \frac{T_{2}}{T_{1}}+\frac{M q_{i c e}}{T_{1}} \\ &\text { where } \quad M q_{i c e}=m\left(q_{2}+c\left(T_{2}-T_{1}\right)\right) \\ &=m q_{2}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)+m c\left(\frac{T_{2}}{T_{1}}-1-\frac{T_{2}}{T_{1}}\right) \\ &=0.2245+0.2564 \sim 0.48 \mathrm{~J} / \mathrm{K} \end{aligned} $$
Phase Transformations
Recommended Textbooks
Elementary Statistics a Step by Step Approach
The Practice of Statistics for AP
Introductory Statistics
Watch the video solution with this free unlock.
EMAIL
PASSWORD