Intermediate Thermodynamics Questions & Answers
 Question by Student 201027111 ( find heat needed to go from b2 to b3) prof said 1st law thermo can't use because unknown veriable is two with pdv and Q but in process V is constant so $dv = 0$ then $d(me) = Q$ then intergral and we can get $me_{b3} - me_{b2} = Q$ and assume calorically perfect gas $e = C_v T$ so $Q = m C_v ( T_{b3} - T_{b2} )$ we can get solve this way is wrong?
 03.19.14
Please check your notation again: $dv$ is not the same as $dV$, b2 is not the same as B2, $p$ is not the same as $P$, $C_v$ is not the same as $C_V$, $me_{\rm b2}$ is not the same as $(me)_{\rm B2}$, etc. Edit your post and use the same notation as used in class, and I will answer your question.
 Question by Student 201312171 Professor, I'd made a lot effort to find the way to reach equation,$\overline {q_x} = \frac {1}{4} \overline {q}$, but couldn't make simple way by integrating vectors on spherical coordinate. However, I found some other method that is to use mathmatical expression on molecules' velocity. M is molar mass, $\bar{R}$ is universal gas constant, T is temperature of gas. In , there is proportional relation $f(q)$ that some particles have velocity $q$ among total particles, a kind of probability function about $q$. Its unit is 'per speed($\frac {1} {\frac {m}{s}}$)' $$f(q) = \sqrt{\left(\frac{M}{2 \pi \bar{R}T}\right)^3}\, 4\pi q^2 e^{- \frac{Mq^2}{2\bar{R} T}}$$ To calcultate average value of $q$, we should multiply it by $q$ and then integrate it in interval $(0,\infty)$. (because $f(q)$ is just a probability function about $q$) $$\int_{0}^{\infty} qf(q)\,{\rm d}q = \int_{0}^{\infty} \sqrt{\left(\frac{M}{2 \pi \bar{R}T}\right)^3}\, 4\pi q^3 e^{- \frac{Mq^2}{2\bar{R}T}}\,{\rm d}q = \sqrt[2]{\frac{8\bar{R}T}{\pi M}}$$ And there is a proportional relation for having 1-D velocity on 'Maxwell-Bolztmann distribution'. If some particle are moving along x-axis direction on xyz coordinate, then the relation is $$f(q_x) = \sqrt{\frac{M}{2 \pi \bar{R}T}} \exp \left[ \frac{-Mq_x^2}{2\bar{R}T} \right]$$ Its unit is also 'per speed($\frac {1} {\frac {m}{s}}$).' To calculate average value of $q_x$, we should multiply it by $q_x$ and then integrate it in interval $(0,\infty)$. $$\int_{0}^{\infty} q_x f(q_x)\,{\rm d}q_x = \int_{0}^{\infty} q_x \sqrt{\frac{M}{2 \pi \bar{R} T}} \exp \left[ \frac{-Mq_x^2}{2\bar{R}T} \right]\,{\rm d}q_x = \sqrt[2]{\frac{\bar{R} T}{2\pi M}}$$ Therefore, comparing two average values of $q$ and $q_x$, there is the relation we've been to figure out. $$\bar{q_x} = \frac{1}{4} \bar{q}$$ I think I should be more hard to make progress on calculating on spherical coordinate. Thanks for reading. Addition : I'm so sorry that I made many mistakes on it. I modified the things. Addition2 : I'm so sorry, sir... I don't know why its unit is wrong with that form... I get the unit that I mentioned, 'per speed'... Did I miss something...?? Addition3 : I'm really really ashamed because I confused 'molar mass' with 'molecular mass'. I really apologize sincerely...=( Reference : wikipedia(Maxwell-Boltzmann distribution, http://en.wikipedia.org/wiki/Maxwell-Bo ... stribution)
 03.21.14
I don't get units of s/m for $f(q)$ as you mention. As long as the units are not consistent, I cannot give you any point for your explanation.. Please double check and edit your post. Only after the units work out will you get a bonus boost. Also get rid of the remaining line breaks. Line breaks should not be used here (they should only be used to separate paragraphs, and there's no need to have more than 1 paragraph per question/comment).
 03.22.14
 Question by Student 201127134 Professor, according to your lecture, there are four types of particles which can carry energy : Molecule, Atom, Electron, Photon. However, Photon, I cannot understand Photon can carry energy. I think "Photon" is just light with no mass. I think, to carry energy, particle must have mass to carry energy, but i think Photon has no mass. How can Photon can carry energy? Ofcourse, you mentioned that there are controversies wheter photon can carry energy. Would you explain how Photon can carry energy?
 04.08.14
This is beyond the scope of this course, and I am not specialized in that field of engineering/physics. But I can give a quick answer. Despite being massless, particules such as photons that travel at the speed of light do carry energy following $E=h c \lambda^{-1}$ with $h$ the Planck constant, $c$ the speed of light, and $\lambda$ the frequency. I can give you 1.0 bonus boost for this question.
 Question by Student 201127146 Professor, I want to ask a question abot perfect gas. To derive specific heat ratio we assumed callorlly and thly perfect gas. I wonder specific heat ratio can be used for non isobaric or non isochoric processes. Similarly, can I assume thermally perfect gas for heating process?
 04.12.14
Yes we can assume thermally perfect gas or calorically perfect gas for any type of thermodynamic process. I'll give you 0.5 point bonus boost for this question. I would have given 1 point if you would have been more careful in typing your question: there are 3 obvious spelling errors ;)
 Question by Student 201127146 I appreciate for your quick response. But I got another questions for assignment. In Q.4. I need to determine specific heat of water. I did my best, but it was out of my range. Can you gibe me a hint to get the value? Next, due to the lecture, entropy is associated with several bodies in one system. No mention about entropy for a body. I wonder if entropy change in 'a body' can have value under 0 Thank you for reading.
 04.13.14
You can find the specific heat of liquid water or gaseous water in the tables — look through them carefully.. And yes, entropy can have a value less than zero for a body that is not isolated from its environment, this was mentioned in the last lecture.. I can not give you any bonus boost for these questions since the answers were given in class already or could be found easily.
 Question by Student 200927102 I have a question that similar with above 201127146's question. I found specific heat of water from your table. There are different values of 3 case. $$i) P=1atm, T=273K, c=4.217kJ/kg\cdot K$$ $$ii) P=1atm, T=300K, c=4.179kJ/kg\cdot K$$ $$iii) P=1atm, T=373K, c=4.218kJ/kg\cdot K$$ These states have pressure condition, but Q.4 in assginment 5, you don't mention about pressure. So I wonder that we must assume that $P=1atm$ state before solve the problem.
 04.14.14
The heat capacity (either for liquids or for gases) doesn't depend much on pressure. You can assume it to be pressure independent. I'll give you 1 point bonus boost for this question.
 Question by Student 201127102 I think your question. Temperature of air is go up, and volume is up too. So air push the piston to argon, and argon's pressure is up. Because piston that between argon and air is insulated, no heat transfer to argon directly. Argon is just given energy from pressure by piston. So, Q to air side is 13429 J, but Q of Argon side is just 1400 J.
 04.16.14
This explanation is only half of the reason behind the high discrepancy between the argon and air raise in energy: there is an additional physical mechanism at play that you are not mentioning.. Think about it more ;) For this half answer, I will give you 1.5 point bonus boost.
 Question by Student 200927143 This question is that you asked about Air and Argon in assignment #2. Times during the course of the actual amount of energy is always conserved(1st Law of Thermo). However, the energy is to be poor quality(2nd Law of Thermo). Quality degradation by increasing the molecular disorder is always accompanied by an increase in entropy. Thermal energy forms that are not inherently organized. Entropy can be viewed as a measure of disorder or disorganization in a system. From the outcome of the problem we can find the entropy change. In Air cylinder entropy change, $$\Delta S = 0.034kg \times (1004J/kg· K \ln\frac{785K}{293K} - 287J/kg· K \times \ln\frac{2atm}{1atm}) = 26.878J/K$$ In Argon cylinder entropy change, $$\Delta S = 0.047kg \times (524J/kg· K \ln\frac{387K}{293K} - 208J/kg· K \times \ln\frac{2atm}{1atm}) = 0.077J/K$$ The entropy change of Argon is smaller than Air. Because of isentropic process in Argon cylinder area. So disorganization in Argon cylinder is smaller than Air cylinder. Therefore thermal energy is less transferred in Argon than Air.
 04.25.14
What you say is not necessarily correct: the rise in temperature of a gas is not always due to a rise in entropy.. Nonetheless, it's a good effort on your part and I'll give you 1.0 point bonus. As I mentioned previously, there is a reason why the air temperature goes up more than the argon that has not yet been outlined.. Think about it more ;)
 04.27.14
 Question by Student 201127134 According to your lecture about Non-Ideal models, intermolecular forces occur between protons and electrons. This is because, electrons are negatively charged, protons are positively charged. However, i think, to make this phenomenon possible, the orbits of each molecules which consists of protons and electrons must be same to make strong attractions that come from close distance between each other. Then, how about the orbits of each molecules are different? I mean, if close distance makes repulsions instead of attractions? Is it still possible to apply Van Der waals equation? I guess, to solve this problem, more GENERALIZED type of hypothesis have to be applied. Is it right?
 04.30.14
What you are asking here is whether the Van der Waals attraction forces would still take place within a mixture of gases (gas with different types of molecules) and not just in a simple gas (gas with one type of molecules). This is a good question. For a mixture of gases, molecules may be next to molecules of the same kind or next to molecules of a different kind. Either way, the fluctuating dipoles will have a tendency to induce other dipoles in other molecules, resulting in groups of molecules spinning in tandem with each other hence resulting in attraction forces. Keep in mind that even within a simple gas, the molecules don't all spin together in a perfectly synchronized way — I explained it this way in class to make it easy to visualize the physical process. Actually, molecules will sometimes be non-synchronized due to them moving around all over the place in random fashion and bouncing off each other. But, many molecules have a tendency to spin in tandem hence resulting in Van der Waals forces even though many more are not synchronized. I liked this question: I'll give you 2 points bonus boost. Have a nice Buddha's birthday holiday :)
 Question by Student 201127101 In lecture, Professor found a and b. $$P_{c}=\frac{\overline{R} T_{c}}{\overline{{v}_{c}}-b}-\frac{a}{\overline{{v}_{c}}^2}$$ $$(\frac{\partial P}{\partial \overline{v}})_{T} = -\frac{\overline{R}T_{c}}{(\overline{{v}_{c}}-b)^2}+\frac{2a}{\overline{{v}_{c}}^3}=0$$ $$(\frac{\partial^{2} P}{\partial \overline{v}^2})_{T} = \frac{2\overline{R}T_{c}}{(\overline{{v}_{c}}-b)^3}-\frac{6a}{\overline{{v}_{c}}^4}=0$$ Solve them, $$∴a=\frac{27\overline{R}^2 T_{c}^2}{64P_{c}}, b=\frac{\overline{R}T_{c}}{8P_{c}}$$ But professor said, the following equation is established at critical point. $$(\frac{\partial^{n} P}{\partial \overline{v}^{n}})_{T} = 0(n = 1, 2, 3 \cdots)$$ So I tried to find a and b different way. $$P_{c}=\frac{\overline{R} T_{c}}{\overline{{v}_{c}}-b}-\frac{a}{\overline{{v}_{c}}^2}$$ $$(\frac{\partial^{2} P}{\partial \overline{v}^{2}})_{T} = \frac{2\overline{R}T_{c}}{(\overline{{v}_{c}}-b)^3}-\frac{6a}{\overline{{v}_{c}}^4}=0$$ $$(\frac{\partial^{3} P}{\partial \overline{v}^{3}})_{T} = -\frac{6\overline{R}T_{c}}{(\overline{{v}_{c}}-b)^4}+\frac{24a}{\overline{{v}_{c}}^5}=0$$ Solve them, $$∴a=\frac{2816\overline{R}^2 T_{c}^2}{6561P_{c}}, b=\frac{11\overline{R} T_{c}}{81P_{c}}$$ I don't know why a and b are different.
OK, I understand your question now, and it's a very good question. Essentially, if you set $\partial ^3 P/\partial \bar{v}^3=0$ and $\partial^2 P/\partial \bar{v}^2=0$ to find $a$ and $b$, then you obtain a different answer as obtained in class. This is correct, you didn't make a mistake. What is happening is that your approach does not guarantee that $\partial P/\partial \bar{v}=0$. In fact, following your approach $\partial P/\partial \bar{v}$ may have any value, either positive or negative! This is not what is happening physically. Thus, we are solving a different problem: I set the first derivative to zero, while you don't, and hence we obtain a different answer. In short, you can only set the higher order derivatives to zero if the lower-order derivatives have already been set to zero. I liked this question, I'll give you 1.5 point bonus boost. I would have given more if you would have gotten the notation right the first time.. ;) Have a nice weekend :)
 05.01.14
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