# Consider two insulated chambers (A,B) of same volume connected by a closed knob, S. One mole of perfect gas is confined in chamber A. What is the change in entropy of gas when knob S is opened? ( R= 8.31 )

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I can't understand for which process the entropy has to be calculated. The question says insulated chamber which indicates that it is an adiabatic process but entropy is constant for an adiabatic process.

Please explain what to do.

Thanks.

I can't understand for which process the entropy has to be calculated. The question says insulated chamber which indicates that it is an adiabatic process but entropy is constant for an adiabatic process.

Please explain what to do.

Thanks.

##### 1 Answer

An **isentropic** (

**adiabatic**(ie no transfer of heat or matter between system and surroundings ... so this is adiabatic)

**and**

**reversible**#color(red)(bblarr) # the**key**idea here

Your intuition should tell you that the gas will **spontaneously** flow through the opened knob and occupy both chambers A & B. That **spontaneity** is synonymous with **irreversibility**, and the arrow of time.

To illustrate, as the system contains 1 mole of perfect gas, the probability of every particle being in chamber

#(1/2)^(N_A) ~~ 10^(-10^23.3)#

It is possible but, in practical terms, impossible - without outside interference.

The system can be returned to its original state... using a reversible process ... that involves work being done on the gas, and heat being released from the gas to its surroundings to keep the temperature constant, quasi-statically, we can even use the thermodynamic definition of entropy.

But the quicker way to explore change in entropy is to split the volume

#N = V/(dV)#

The number

#Omega = ""^NP_n qquad "where "{(n = "no of particles of gas"), (""^NP_n = (N!)/((N-n)!) ):}#

Doubling the available volume means:

From Boltzman's definition of entropy:

#S = k_B ln Omega#

The change in entropy is:

For one mole:

So :

#Delta S = R ln 2#

You can construct a reversible process that takes the system from its initial to final state, or vice versa, but that process is no longer adiabatic.

Final thought:

# {( dS= (dQ_(color(red)(rev)))/T), ( dS ge (dQ )/T):}#