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I have attempt the question.Howeve i had find out some contradiction but i could not understand where is the mistake i made.Below is my working and the answer given by the book A cylinder is close at both ends and has insulating walls. It is divided into two compartments by a perfectly insulating partition that is perpendicular to the axis of the cylinder. Each compartment contains 1.00 mol of oxygen, which behaves as an ideal gas with γ=7/5. Initially the two compartments have equals volumes, and their temperatures are 550k and 250k. The partition is then allowed to move slowly until the pressure on its two sides are equal. Find the final temperatures in the two compartments. Let P 1 be the initial pressure in 1 st compartment P 2 be the final pressure in 1 st compartment P 3 be the initial pressure n 2 nd compartment P 4 be the final pressure in 2 nd compartment V 1 be the initial volume in 1 st compartment V 3 be the initial volume in 2 nd compartment V2 be the final volume in 1 st compartment V4 be the final volume in 2 nd compartment T 1 be the initial temperature in 1 st compartment T 2 be the final temperature in 1 st compartment T 3 be the initial temperature in 2 nd compartment T 4 be the final temperature in 2 nd compartment X be the length of the cylinder Y is the distance move by the partition and is an unknown which must be find and in the range from 0 to 0.5 Solving the equation 1 and 2 P 1 V1 γ =P 2 V 2 γ -1 P 3 V3 γ =P 4 V 4 γ -2 Will get the ration V 2 =1.756V 4 or T 2 =1.756T 4 Than consider the length of the whole compartment be X since initially two compartment same volume, the partition divide the cylinder into 0.5X at both side and consider the distance move by the cylinder be YX. since the partition will move to the right, the volume of the 1 st compartment when the partition stop is (0.5X+YX)A where A is the cross section of the partition and for the 2 nd compartment will be (0.5X-YX)A solve (0.5X+YX)A=1.756(0.5X-YX)A will get Y=0.137 and use T 1 V 1 γ-1 =T 2 V 2 γ-1 will get T 2 =499K and T 4 =284K Let ?T=T 2 +T 4 -800 Differentiate this equation T 1 (0.5XA/(0.5+Y)XA)^0.4+T 2 (0.5XA/(0.5-Y)XA)^0.4-800=?T When d(?T)/dT=0 Y=0.137 But answer given Since the work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compress gas. nR/γ-1(T 1 -T 3 )=-nR/γ-1(T 4 -T 2 ) T 2 +T 4 =T 1 +T 3 =800K And substitute T2=1.756T4 into the equation will get T 2 =510K and T 4 =290K. The distance move by the partition until it stop, should be consider this way. T 1 (0.5X/(0.5+Y)X)^0.4+T 2 (0.5X/(0.5-Y)X)^0.4-800=?T Partition stop when ?T=0 And solve the equation T 1 (0.5X/(0.5+Y)X)^0.4+T 2 (0.5X/(0.5-Y)X)^0.4-800=0 To get the value of Y Y=0.258 At Y=0.137 where the pressure on both side is the same, T 2 +T 4 =784K but not 800K Y T 2 +T 4 0.000 800.00 0.050 790.18 0.100 784.65 0.130 783.43 0.137 783.38 0.150 783.54 0.200 787.41 0.250 797.57 0.258 800.00 I had discuss the problem with several teacher but it seem that none of the answer i feel like making sense. At 1st, i thought it may be due to the cylinder is not a uniform cylinder that why the formula cannot be apply. On 2nd thought, the adiabatic formula should be true for all form no wonder it is uniform or not uniform. secondly, it can be due to approximation error but i have use scientific key in the formula calculate as accurate as possible still the value remain different. If it is cause by approximation formula can someone prrof it? Since previously question answered that translational kinetic energy and kinetic energy is different, could the gas particle convert portion of its translational energy to kinetic energy so there is a net movement of gas particle toward the right side? causing the total internal energy in the system appear different(there is a "loss" of internal energy) because the translational kinetic energy is use to do work on the right partition and increase in kinetic energy?From the formula i find that the "loss" in internal energy is maximum when both side is same pressure.


I have attempt the question.Howeve i had find out some contradiction but i could not understand where is the mistake i made.Below is my working and the answer given by the book


A cylinder is close at both ends and has insulating walls. It is divided into two compartments by a perfectly insulating partition that is perpendicular to the axis of the cylinder. Each compartment contains 1.00 mol of oxygen, which behaves as an ideal gas with γ=7/5. Initially the two compartments have equals volumes, and their temperatures are 550k and 250k. The partition is then allowed to move slowly until the pressure on its two sides are equal. Find the final temperatures in the two compartments.


 


Let P1 be the initial pressure in 1st compartment


      P2 be the final pressure in 1st compartment


      P3 be the initial pressure n 2nd compartment


      P4 be the final pressure in 2nd compartment


      V1 be the initial volume in 1st compartment


      V3 be the initial volume in 2nd compartment


      V2 be the final volume in 1st  compartment


      V4 be the final volume in 2nd compartment


      T1 be the initial temperature in 1st compartment


      T2 be the final temperature in 1st compartment


      T3 be the initial temperature in 2nd compartment


      T4 be the final temperature in 2nd compartment


       X be the length of the cylinder


       Y is the distance move by the partition and is an unknown which must be find and in    


       the range from 0 to 0.5


 


Solving the equation 1 and 2


P1V1γ=P2V2γ -1


P3V3γ=P4V4γ-2


 


Will get the ration V2=1.756V4 or T2=1.756T4


Than consider the length of the whole compartment be X since initially two compartment same volume, the partition divide the cylinder into 0.5X at both side and consider the distance move by the cylinder be YX. since the partition will move to the right, the volume of the 1st compartment when the partition stop is (0.5X+YX)A where A is the cross section of the partition and for the 2nd compartment will be (0.5X-YX)A solve (0.5X+YX)A=1.756(0.5X-YX)A will get Y=0.137 and use T1V1γ-1=T2V2γ-1 will get T2=499K and T4=284K


 


 


Let ?T=T2+T4-800


Differentiate this equation


T1 (0.5XA/(0.5+Y)XA)^0.4+T2(0.5XA/(0.5-Y)XA)^0.4-800=?T


When d(?T)/dT=0


Y=0.137


 


But answer given


Since the work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compress gas.


 


nR/γ-1(T1-T3)=-nR/γ-1(T4-T2)


T2+T4=T1+T3=800K


 


And substitute T2=1.756T4 into the equation will get T2=510K and T4=290K.


 


The distance move by the partition until it stop, should be consider this way.


 


T1 (0.5X/(0.5+Y)X)^0.4+T2(0.5X/(0.5-Y)X)^0.4-800=?T


Partition stop when ?T=0


And solve the equation


T1(0.5X/(0.5+Y)X)^0.4+T2(0.5X/(0.5-Y)X)^0.4-800=0


To get the value of Y


Y=0.258


At Y=0.137 where the pressure on both side is the same, T2+T4=784K but not 800K


 


   Y              T2+T4


0.000           800.00


0.050           790.18


0.100           784.65


0.130           783.43


0.137            783.38


0.150            783.54


0.200            787.41


0.250            797.57


0.258            800.00


 I had discuss the problem with several teacher but it seem that none of the answer i feel like making sense.


At 1st, i thought it may be due to the cylinder is not a uniform cylinder that why the formula cannot be apply. On 2nd thought, the adiabatic formula should be true for all form no wonder it is uniform or not uniform.


secondly, it can be due to approximation error but i have use scientific key in the formula calculate as accurate as possible still the value remain different. If it is cause by approximation formula can someone prrof it?


Since previously question answered that translational kinetic energy and kinetic energy is different, could the gas particle convert portion of its translational energy to kinetic energy so there is a net movement of gas particle toward the right side? causing the total internal energy in the system appear different(there is a "loss" of internal energy) because the translational kinetic energy is use to do work on the right partition and increase in kinetic energy?From the formula i find that the "loss" in internal energy is maximum when both side is same pressure.

Grade:Upto college level

1 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

There is an error in the differentiation part.Please check again...

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