Question

A piston can slide without friction inside a horizontal cylindrical vessel,which contains an ideal mono atomic gas.The piston and the cylinder both are made of a perfect heat insulating material. Initially the piston is in equilibrium and divides the cylinder into two parts A and B,which are not necessarily equal.The temperature of the gases in both the parts are equal.Now,the piston is held in its initial position and the gas in part A is supplied some amount of heat,then the piston is released.What will the piston do in its subsequent motion.

Answer 1

@ganeshie8 @Michele_laino @parthkohli

Answer 2

Sir I made some equations..
Shall I inform you about the same??

Answer 3

yes! thanks!

Answer 4

Firstly we know that moles will remain unchanged in part A and part B both so
First is like
Pa*r/Ta=PA*(r+x)/Ta'
Here r is the distance of the piston from extreme (left) n I assumed the whole length of container as L
X is the displacement of piston after supply of heat

Answer 5

Away from gas in part A...

Answer 6

2nd is Pb*(L-r)/Ta=Pb'(L-r-x)/Tb'

Answer 7

I took temperature in the first condition of vessel to be Ta

Answer 8

I ask, why do you think that there will be an expansion?

Answer 9

Heat is supplied to the part A that's why I thought so??
Am I wrong in my approach?

Answer 10

I made that question, since it is what the exercise is asking

Answer 11

Since it's an adiabatic case so v can say that the internal energy would be providing the necessary work for expansion

Answer 12

ok! yes, that's right!
here is the equation for an adiabatic trasformation:
\[\Large {P^{1 - k}}{T^k} = {\text{costant}}\]

Answer 13

Sir I got this now...
Pa*r^y=Pa'*(r+x)^y
And Pa*(L-r)^y=Pb'*(L-r-x)^y
And y stands for the adiabatic constant(gamma)

Answer 14

where:
\(k=C_p/C_v\)
please wait, from the first principle of thermodynamic, we have this:
\[\Large \Delta U = - L > 0 \Rightarrow L < 0\]
in other words, the work done by th fluid A has to be negative

Answer 15

thermodynamics*

Answer 16

Sir i applied this concept
PV^y =k

Answer 17

yes! It is the same

Answer 18

From that i got those two equation which i wrote just above ..

Answer 19

I saw your equations, please comment my last statement

Answer 20

And furthermore v can say that
(Pa'-Pb')A=ma

Answer 21

I didn't see yours at the first attempt..!
And yes sir they r definitely the same!!

Answer 22

M is mass of piston

Answer 23

please, from my last statement:
\(L<0\)
we can say that fluid A is doing a negative work, namely its volume will decrease, am I right?

Answer 24

Yep! It is ...
Well i didn't give it a damn becoz what we r concerned here with is the motion of mass m ...so we can say that it will not change much to our ans

Answer 25

But if you wish the same then i can repicturise these equation..

Answer 26

I think that there is something wrong.
Namely
we can say that the fulid A is doing an expansion, then the first principle,
\[\Delta U = - L > 0 \Rightarrow L < 0\]
can be applied for fluid B only, so we can say that fluid B is doing a contraction
and your equations are right!

Answer 27

Sir bt i m unable to make a shm type equation from those two particular equations...
Will u pls help in that portion?

Answer 28

qualitatively spaking, we have solved the question using the first principle of thermodynamics:
\[\large Q = \Delta {U_A} + {L_A},\quad {L_A} > 0\]
for part A, and:
\[\Large \Delta {U_B} = - {L_B} > 0 \Rightarrow {L_B} < 0\]
for part B, since fluid inside B undergoes to an adiabatic compression

Answer 29

and the temperature of fluid B will increase

Answer 30

U mean to say that temperature in A will decrease! And that in B increase

Answer 31

Sir finally i got theses..
Pa[r^y(r+yx)-(L-r)^y(L-r+yx)]=ma

Answer 32

I mean that the temperature of B only, will increase

Answer 33

These*

Answer 34

do you want to write an equation of motion for piston ?

Answer 35

Sorry i made a mistake in very first part it should be r^y(r-yx)...
Yes sir!!

Answer 36

please, do you know what is the type of motion of the piston?

Answer 37

I will know it after knowing the equation.

Answer 38

I think that it can be an oscillatory motion

Answer 39

If it comes out to be a is directly proportional to x then it is shm

Answer 40

How will v prove it??
I told the final equations that i came up with ..
After that i m not getting ny way to solve them..
Will u like to help in these?

Answer 41

it is not an easy question, since there are many variables, namely temperatures of A and B, pressures of A and B

Answer 42

Sir bt the equation i made finally was free of all the variables and only there were Pa, r ,L and x which is great thing to solve the equation. .

Answer 43

The only thing is v have to bring in terms of x n then v r done!!

Answer 44

we start to give a little displacement to the piston, therefore we have to write the forces which are acting on such pistons.
all what we know, is that after the heat exchage, we have all adiabatic trasformation of A and B fluids

Answer 45

piston*

Answer 46

I gave that x was the slight displacement only...

Answer 47

please wait, I'm writing some equations

Answer 48

after a little displacement \(x\) to the right, we get the new values of volume and temperature, as function of the old ones, like below:
\[\Large \begin{gathered}
{V_A}' = {V_A} + Sx \Rightarrow {P_A}' = {P_A}{\left( {\frac{{{V_A}}}{{{V_A} + Sx}}} \right)^k} \hfill \\
\hfill \\
{V_B}' = {V_B} - Sx \Rightarrow {P_B}' = {P_B}{\left( {\frac{{{V_B}}}{{{V_B} - Sx}}} \right)^k} \hfill \\
\end{gathered} \]
where \(S\) is the cross sectional area of the piston.

Answer 49

This Va/Va+Sx can be even simplified to r/r+x...

Answer 50

so we can write this differential equation:
\[\Large M\ddot x = \left\{ {{P_A}{{\left( {\frac{{{V_A}}}{{{V_A} + Sx}}} \right)}^k} - {P_B}{{\left( {\frac{{{V_B}}}{{{V_B} - Sx}}} \right)}^k}} \right\}S\]
where \(M\) is the mass of the piston

Answer 51

r is the initial length of part A

Answer 52

Correct sir!!

Answer 53

ok!

Answer 54

now, as we can see, such ODE is not a \(linear\) ODE, so we have to linearize it

Answer 55

By binomial ...ryt?? I applied that even..after that only i got these..
Pa[r^y(r-yx)-(1-r^y)(L-r+yx)]=ma

Answer 56

\[\Large M\ddot x = \left\{ {{P_A}{{\left( {\frac{r}{{r + x}}} \right)}^k} - {P_B}{{\left( {\frac{{d - r}}{{d - r - x}}} \right)}^k}} \right\}S\]

Answer 57

Sure sir...these are in match vid my equation

Answer 58

Now if v bring that denominator term in numerator side v can solve them using binomial..

Answer 59

I think that we can write the Taylor's expansion at the first order, for both this functions:
\[\Large {\left( {\frac{r}{{r + x}}} \right)^k},\quad {\left( {\frac{{d - r}}{{d - r - x}}} \right)^k}\]