One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM

DETAILS

MRP

DISCOUNT

FINAL PRICE

Total Price: Rs.

There are no items in this cart.

Continue Shopping

Continue Shopping

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.

10 years ago

Dear student,

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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

**Win ****exciting gifts**** by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.**

**Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : ****Click here**** to download the toolbar..**

**Askiitians Expert**

**Sagar Singh**

**B.Tech, IIT Delhi**

copyright © 2006-2021 askIITians.com

info@askiitians.com