Q. 225.0( 2 Votes )

An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5 nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find

(a) the work done by the gas in the left part during the process.

(b) the temperature on the two sides in the beginning,

(c) the final common temperature reached by the gases,

(d) the heat given to the gas in the right part and

(e) the increase in the internal energy of the gas in the left part.

Answer :

a) According to the question, conducting separator between both the part is fixed. So, no change in volume will be observed. And as we know work done a gas is PV. Therefore, no work will be done on the left part during the process as the volume is not changing.


b) Given


The vessel is divided into two equal parts.


So, the volume of left part V1 and volume of right part V2 will half of the total volume.



Initial pressure on each side is p


Number of moles on left side n1 =1


Number of moles on right side n2=2


Let initial temperature for the left and right part be T1 and T2 respectively.


For the left part, applying the ideal gas equation




Similarly, for the right part




c) It is given that the internal energy of the gas is


U=1.5nRT


Where n=number of moles


R=gas constant


T=final equilibrium temperature of both parts


The internal energy of ideal gas is given as


U=nCvT


Where Cv=molar specific heat at constant volume


T=temperature.


Total moles n=n1+n2=1+2=3


U=3CvT


Let U1 and U2 be the internal energy of the left and right part respectively.


So, U1=n1CvT1= CvT1


And U2=n2CvT2=2CvT2


(gas is same on both the part so Cv will be same)


Now


U=U1+U2


3CvT = CvT1+ 2CvT2


3T=T1+2T2




d) As stated in part (a) no work will be done on either part of the vessel as conducting separator is fixed.


So, ΔW=0 for the right part of the vessel.


From first law of thermodynamics, we know that,


ΔQ=ΔU+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


ΔQ=ΔU


It is given that the internal energy of the gas is


U=1.5nRT


Where n=number of moles


R=gas constant


T=final equilibrium temperature of both parts


ΔU=1.5nRΔT


For the right part, change in internal energy after equilibrium has reached will be due to the change in temperature from T2 to T.


ΔU=1.5n2R(T-T2)





Thus, heat given to the right part is .


e) Since ΔW is zero as the volume is fixed, therefore, the first law of thermodynamics.


But since heat is given to the right part that means heat is extracted from the left part. So, the internal energy of the left part of will decrease.


Therefore, heat given to the right part will be equal to the negative internal energy of the left part.


ΔQ=-ΔU (left part)


.


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