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
@mayankdevnani pls..help..
@imqwerty wud u help?
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
Note that we can write this only under the assumption that both these gases do not react with each other.
How are u able to find out the final moles...?
@parthkohli pls...help..
Sorry bt i didn't get from where u came to the formula of energy of gas as ynRT
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 \(\gamma\).
Okkk...
I thought it was gamma
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.\[\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2} = \frac{P(V_1+ V_2)}{T}\]
I almost forgot that \(\gamma\) was a reserved symbol. Sorry about that, haha.
Actually one more question here the term equilibrium implies temp. equilibrium Or pressure equality
Or probably both? Equilibrium just means that the system just "settles" down so that there is no further change in the state variables.
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
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
Correct.
Tysm parth ....it meant to be a gud concept to me ...ur method...
No problem. :) What answer do you get?
The same u came up vid...
Oh, BTW, you'll have to use both equations.
Oh sorry....i forgot it completely... I were thinking dat ans came jst by the use of gaseous state equation... Sorry for dat
Okk....dis was d best approach of dis question....thank u once again for giving ur precious tym to dis question
=)
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