Question

An ideal gas with energy U and N atoms is held in one-half of a chamber of volume V by a partition (refer Figure 1(A)). As the partition is removed and the gas in the container is allowed to occupy the entire volume V, an additional N atoms of the same gas with the same energy U is introduced into the container as shown in Figure 1(B). . Nu, V/2A) Figure 1 1. What is the initial entropy (S) of the gas (before removing the partition and adding an additional N atoms of the gas)?

Answer 1

entropy s of an ideal gas is given by is! S = Nk len v 1 417 ņu +5 IN I3NK L ha Planck's keboltzmann constant. constant mis mass of an atom : - . Vi a initial volume a v 1. initial entropy = 8 1.Siinklin (V/2 (4.A mŲ 192|15| TINH² Vi final volume = v We a number of atoms in the final state = 2N :- final inete internal energy 20

z final entropy Sp = No k l m ( N (ARM) In Vf 14am us 1 I N 1 3 Ngh s - 2NKU V 14km zu 1/1 N 13X ZNK? 2 = 2NK Ten/v 8amu 12 + 5 INI GNK / 1

I 2N (V (4kmu/ . 2N | SNAI 22 Sé Ang as Sf - Sj. = 25; Si = Se

This does not ho tell us much about the change in entropy since si itself to looks very complicated Another way to look at this process is to say that the given process is essentially a mining process where 2 N number of atoms mix in volume V. - we know that change in entropy. of in the process of mining AS в духи by TAS - aNR 12, en xit ų en al 1 Where nt no. of moles of gas L x, mole fraction of fisst a component X mole fraction of second. S. Componenti Tin our case na Nittaa n = 2Nb ayagadrós DNA A number t2 = - 2 = = N 2N. ZN 21 IN

Thence as = - 2NR I ent tenta =- 2NK in /= ankln2.