how can we separate two gases which r mixed together??
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No, it's not possible. but we can separate two solid or liquid things .
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Heya!!!
No
Because you aren't told, there are three possible cases to consider (well there's more, but these are the natural cases):
Two monoatomic GasesOne monoatomic and one diatomic gasTwo diatomic gases
We should just note that the equilibrium for this system will be the state at which the temperature of each system is equal. We want to find where that temperature ends up between T1T1 and T2T2. Let's call this TfTf. So we have,
T1≤Tf≤T2.T1≤Tf≤T2.
Also note that this a constant volume process not a constant pressure one.
Two Monoatomic Gases:Two Monoatomic Gases:
The heat capacity of a monoatomic ideal gas is simply Cv=32nRCv=32nR.
Then, the internal energy of the system at a given temperature is,
U=32nRT.U=32nRT.
Then, the change in internal energy of one system is the negative of the change in internal energy of the other,
ΔU1=−ΔU2ΔU1=−ΔU2
so that,
32nR(Tf−T1)=32nR(T2−Tf)32nR(Tf−T1)=32nR(T2−Tf)
Upon rearranging this yields,
Tf=T2+T12Tf=T2+T12
which gives the (perhaps) intuitive result that the final temperature the gases is simply the average of the two initial temperatures for the fairly restrictive conditions we have here.
No
Because you aren't told, there are three possible cases to consider (well there's more, but these are the natural cases):
Two monoatomic GasesOne monoatomic and one diatomic gasTwo diatomic gases
We should just note that the equilibrium for this system will be the state at which the temperature of each system is equal. We want to find where that temperature ends up between T1T1 and T2T2. Let's call this TfTf. So we have,
T1≤Tf≤T2.T1≤Tf≤T2.
Also note that this a constant volume process not a constant pressure one.
Two Monoatomic Gases:Two Monoatomic Gases:
The heat capacity of a monoatomic ideal gas is simply Cv=32nRCv=32nR.
Then, the internal energy of the system at a given temperature is,
U=32nRT.U=32nRT.
Then, the change in internal energy of one system is the negative of the change in internal energy of the other,
ΔU1=−ΔU2ΔU1=−ΔU2
so that,
32nR(Tf−T1)=32nR(T2−Tf)32nR(Tf−T1)=32nR(T2−Tf)
Upon rearranging this yields,
Tf=T2+T12Tf=T2+T12
which gives the (perhaps) intuitive result that the final temperature the gases is simply the average of the two initial temperatures for the fairly restrictive conditions we have here.
akashdas95:
dear its too complicated plz explain in short
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