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Physics 17 Online
OpenStudy (samigupta8):

Two thermally insulated vessels 1 n 2 are filled vid air at temperatures (T1,T2) , volume (V1,V2) and pressure (P1,P2) respectively.If the valve joining the two vessels is opened the temperature inside the vessel at equilibrium is

OpenStudy (samigupta8):

@mayankdevnani pls..help..

OpenStudy (samigupta8):

@imqwerty wud u help?

OpenStudy (aravindg):

The resultant number of moles will be sum of number of moles of each gases. Use n=pv/RT . Use that for initial case and final case. nA+nB=nfinal

OpenStudy (aravindg):

Note that we can write this only under the assumption that both these gases do not react with each other.

OpenStudy (samigupta8):

How are u able to find out the final moles...?

OpenStudy (samigupta8):

@parthkohli pls...help..

OpenStudy (samigupta8):

Sorry bt i didn't get from where u came to the formula of energy of gas as ynRT

Parth (parthkohli):

I'm calling \(\gamma\) to be\[\gamma = \frac{f}2\]where \(f\) is the number of degrees of freedom of the gas. I'm sorry if it was mistaken to be \(\gamma = C_p/C_v\). Basically what I'm saying is that it doesn't matter what we multiply to the \(nRT\) term... it'll cancel out. I called that constant \(\gamma\).

OpenStudy (samigupta8):

Okkk...

OpenStudy (samigupta8):

I thought it was gamma

Parth (parthkohli):

Maybe energy conservation can also help, as there is no loss of energy to the environment. We know that the energy of a gas is given by \(\gamma n RT\) (where \(\gamma = f/2)\).\[\gamma \frac{P_1V_1}{RT_1}\cdot RT_1 + \gamma \frac{P_2V_2}{RT_2}\cdot RT_2 = \gamma \frac{P(V_1+V_2)}{RT}\cdot RT\]\[\Rightarrow P_1V_1 + P_2 V_2 = P(V_1 + V_2)\]In addition to the equation we found earlier.\[\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2} = \frac{P(V_1+ V_2)}{T}\]

Parth (parthkohli):

I almost forgot that \(\gamma\) was a reserved symbol. Sorry about that, haha.

OpenStudy (samigupta8):

Actually one more question here the term equilibrium implies temp. equilibrium Or pressure equality

Parth (parthkohli):

Or probably both? Equilibrium just means that the system just "settles" down so that there is no further change in the state variables.

OpenStudy (samigupta8):

Correct then it vud be more convincing result for me ....the result u derived n Evn now d same can b found by applying state of gas equation P1V1/RT1 + P2V2/RT2= P(V1+V2)/RT

OpenStudy (samigupta8):

Bt it has to be d equilibrium of both the states P n T othrwise u can think dat i didn't get it... Lol

Parth (parthkohli):

Correct.

OpenStudy (samigupta8):

Tysm parth ....it meant to be a gud concept to me ...ur method...

Parth (parthkohli):

No problem. :) What answer do you get?

OpenStudy (samigupta8):

The same u came up vid...

Parth (parthkohli):

Oh, BTW, you'll have to use both equations.

OpenStudy (samigupta8):

Oh sorry....i forgot it completely... I were thinking dat ans came jst by the use of gaseous state equation... Sorry for dat

OpenStudy (samigupta8):

Okk....dis was d best approach of dis question....thank u once again for giving ur precious tym to dis question

Parth (parthkohli):

=)

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