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Ia GasMcljty uiithum idinjc90Wmlulistinjtncrcum 4*.FnSnmMtanaqecttriDrdlarut71b79u7z Wamt...

Question

Ia GasMcljty uiithum idinjc90Wmlulistinjtncrcum 4*.FnSnmMtanaqecttriDrdlarut71b79u7z Wamt

Ia Gas Mcljty uiithum idinjc90W mlulistinjtnc rcum 4*. FnSnm Mtan aqecttri Drdlarut 71b79 u7z Wamt



Answers

Let $\left(p_{0}, V_{0}, T_{0}\right)$ be the initial state of the gas. We know $A_{\text {adia }}=\frac{-v R \Delta T}{\gamma-1}$ (work done by the gas) But from the equation $T V^{\gamma-1}=$ constant, we get $\Delta T=T_{0}\left(\eta^{\gamma-1}-1\right)$ Thus $A_{\text {adia }}=\frac{-v R T_{0}\left(\eta^{\gamma-1}-1\right)}{\gamma-1}$ On the other hand, we know $A_{i s o}=v R T_{0} \ln \left(\frac{1}{\eta}\right)=-v R T_{0} \ln \eta$ (work done by the gas) Thus $\quad \frac{A_{\text {adia }}}{A_{\text {iso }}}=\frac{\eta^{\gamma-1}-1}{(\gamma-1) \ln \eta}=\frac{5^{0-4}-1}{0 \cdot 4 \times \ln 5}=1 \cdot 4$

In this problem, I can write the reaction energy H two, C 204 then come in emergence of heat. Then it will give, I guess, be just look at it carefully. The gas B age see all plus yes, C. 02 plus liquid as two oh and gas. Be that did C. O plus CL two will react to give the guests. S and guess passage when react with when gas has react with an S. Three, it will produce N. H two C. O. Manage to as the product. So option B is correct here.

Among the given options, among the given options, among the given options in this problem, among the given options in this problem, The gas 03, the gas rotary cannot oxidized, cannot oxidized GM and no fool P. M. And all for so. According to the option years of some age, correct answer for this problem of some it's correct answer for this problem. I hope you understand the solution of this problem.

Hello students in this question. In the foregoing problem we have to assume that that this car located in the ultra rare field guests of molar mass, capital M at a temperature T. And pressure peaks. Okay, so we have to find this end. Okay. Okay. No, we can stay here that the data for this condition is one by three average speed v average molecular Belinda molecular density through. Okay. And here the lambda will be close to one by two times of edge. Okay, so we can substitute the values of viscosity coefficient. It'll will be won by three more player bait. This average spill it is eight K temperature T divide by pi manipulated by a small M. And multiplayer beta lambda which is one by two times of edge and density wrote this will be must never be pressure developed Katie. Okay. From the ideal gas situation. So this can be rearranged so that we can right Eat up. It is equal to one x 3 times of to capital M divided by pi R temperature T. Molecular bay HP Underwood. Okay. Using the fact that our by N. A. This is equal to Goldsman constant K. And mass murder club and A. It is equal to molar mass. Okay, so using these two equations, we have calculated this value. No, we can calculate this eat and which is the moment of the friction force. So and will be given by from the previous problem one by to watch. Multiply by pi ETa here to the power for market share by omega soap. I get to the powerful omega divided by two edge and molecular by this cheetah, which is this value. So one by three and the root of two capital M. Divided by pi capital artie mark played by ach pressure P. Okay, So after solving this and it is equal to one by three omega air to the powerful pressure P. And the root of pi capital M. The worst day to our temperature. T. So this is the answer for this problem. Okay, thank you.

The question is laughing guest is laugh fingers, Let's start guys with you, Is that responsible for the love of a human being? So formula of loving, this is the end.


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