The change of free energy with pressure for one mole of a perfect gas is at constant temperature is delta G = RT ln(P2/P1)
Is the above statement correct. If yes then how??? If no, then what's the mistake in it??
$G=$Go +RTLOGQ, Q=reaction quotient. For gases,Q is in partial pressure of the gas.In your equation,$Go is missing.($=delta)
Well exactly i also found the same mistake in this eqn. and didn't mark it as correct in one of my exams.... I dont know the correct answer till yet.... but will let u know the same as soon as i get it.... But some of my friends were saying that it is correct as it is given as a standard eqn. in some books... but i myself have not seen it in any book.... Lets see....
Well,i would suggest you to read ncert thermod(last part) .I think the eq given is valid only for reaction and not for pure gas.
Well Chandni, i have received the answers of my test... and this statement has been given as a correct statement. Don't have the solutions yet... so dont know how's it correct... Any comments???
I think equation can be correct if delta Go is zero i.e. reaction attains equilibrium at room temperature.
chandni and ankur , the eq you talkin abt is used whenever a rxn is at equili or may attain , the basic definition of gibbs free energy is given by
G = H - TS
at contant temperature , change in enthalpy is zero , and i hope you can calculate TS at constant temp
Shreyank.... I didn't get u.... can u please show me the whole steps how u r getting this eqn.???
You see all those people are wrong, it's not at all like this .I too got the same doubt when i was studying the Nerst equation.Basically what we have to consider in the logarithm is the reaction quotient & the reaction quotient is not defined as the ratio what we have defined in chemical equilibrium i.e its not the ratio of product of active concentrations ......... but it is the ratio of ACTIVITIES which we see later in our life if we are to do some research in quantum mechanics. So you need not bother much about it .But to say what we write in euilibrium is wrong is not right. As we deal with fairly low concentrations we have activities= concentration, this is the reason why pH isn't -ve.Here aloso we take the -ve logarithm of the ACTIVITY of the H+ ion and we know that activity cant exceed 1.Hence pH ain't -ve.So to say ln{p1/p2} is a real mess,but in our level we are justified to use it but do prefer using concentration, as the ideal gas equation is not valid in our normal chemical systems .You are a smart fellow you should be knowing this.If you say that this is thermodynamics you are mistaken even in thermo dynamics we use activities and not pressures.
MANOJ
MANOJ... Whats ur final answer???
I say in our level it is yes but actually speaking it a BIG NOOOOOOOOOOOOOOOOOOOOOOO
. Read my passage again carefully ihave already mentioned the answer 
. Read my passage again carefully ihave already mentioned the answer
Well u may be right Manoj... But in the answers of my exam, its given that it is correct.
What but? I too am saying that the answer is correct, i was only explaining you why i have done it.If i were writng your exam i would definetly put it as correct.Moreover do you know one thing a certain compound can have two solubility products.It goes like this, you see whn we discuss about the concentrations we put 10^-40 M solution but practically speaking is it possible.well Iwill put it up as a question i do know the answer but lets see what you and _______ got to say about it
Well Manoj,
Your reply is a very sensible one except for the fact that you are being too general by quoting quantum mechanics and stuff.
Can you reword yor reply sticking just to the points?
It would be easier to comprehend.
And please write in paragraphs.
Your reply is a very sensible one except for the fact that you are being too general by quoting quantum mechanics and stuff.
Can you reword yor reply sticking just to the points?
It would be easier to comprehend.
And please write in paragraphs.