Q21 of 51 Page 62

Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides.

(a) How much work has been done by the gas on the left part?


(b) Find the final pressures on the two sides.


(c) Find the final equilibrium temperature.


(d) How much heat has flown from the gas on the right to the gas on the left?



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


First, we will calculate the final temperature and then final pressure.


(c) Given


Pressure of left chamber =P1


Pressure of right chamber =P2


Temperature of left chamber =T1


Temperature of right chamber =T2


Let the number of moles in the left chamber be n1


Number of moles in the right chamber be n2


Diathermic wall has divided the tube in two equal part. So, the volume of the left and the right chamber will be V/2.


Applying ideal gas equation in the left chamber,




Similarly applying ideal gas equation in the right chamber,




Total number of moles n=n1 +n2





The internal energy of ideal gas is given as


U=nCvT


Where Cv=molar specific heat at constant volume


T=temperature.


According to question,


U=1.5nRT


nCvT=1.5nRT


So, Cv=1.5R


The internal energy of the left chamber U1=n1CvT1


The internal energy of right chamber U2= n2CvT2


Total internal energy U= U1+U2


1.5nRT= n1CvT1+ n2CvT2


1.5nRT=Cv(n1T1+n2T2)


Substituting the value of Cv


1.5nRT=1.5R(n1T1+n2T2)


nT=n1T1+n2T2


substituting the value of n1 and n2 in above equation





Substituting the value of n in the above equation,




Thus, final equilibrium temperature T=.


(b) Now we will find final pressures on both sides.


Let final pressure of left chamber P1’


Final pressure of right chamber P2’


Applying ideal gas equation in the left chamber before and after equilibrium




From equation (i) and (ii),




Substituting the value of T,



Applying ideal gas equation in the right chamber before and after equilibrium




From equation (iii) and (iv),




Substituting the value of T,



(d) The internal energy of ideal gas is given as


U=nCvT


Where Cv=molar specific heat at constant volume


T=temperature.


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


So, ΔW=0 for the right chamber of the tube.


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


Change in internal energy of the right chamber after equilibrium has reached will be


ΔU=n2CvT2 -n2CvT


Substituting the value of n2, Cv and T in the above equation








Thus, heat flown from left to right chamber is
.


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Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J kg–1 K–1 and its densities at 0°C and 4°C are 999.9 kg m–3 and 1000 kg m–3 respectively. Atmospheric pressure = 105 Pa.

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Calculate the increase in the internal energy of 10g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg m–3. The specific heat capacity of water = 4200 J kg–1 °C–1 and the latent heat of vaporization of water = 2.25 × 106 J kg–1.

 22

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.