- #1

bbolddaslove

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## Homework Statement

A container is divided into two parts by a thermally conducting wall. There are N atoms of a monatomic ideal gas on the left side, 2N on the right. The gas on the left is initially at absolute temperature 200K, the gas on the right at 500K.

a. After thermal equilibrium is reached, what is the temperature in the box (on either side)?

b. How much does the entropy of the whole system change in reaching thermal equilibrium?

T_L=N

T_R=2N

T_f=final temp

## Homework Equations

E=(3/2)N*k_b*T

deltaE=deltaQ

dS=(k_b*T)^-1*dQ

## The Attempt at a Solution

a. For the first part, I used the equation for E to get (3/2)N*k_b*(T_f-T_L)=deltaQ and (3)N*k_b*(T_f-T_R)=-deltaQ, because deltaE_L=-deltaE_R

(3/2)N*k_b*(T_f-T_L)=-(3)N*k_b*(T_f-T_R), then crossing out k_b and N and solving:

T_f-T_L=2(T_f-T_R)

so T_f=2T_R-T_L = 1000K-200K = 800K

I'm not sure if i did this right or not

b. dS=(k_b*T)^-1*dQ and dQ=dE, so

dS=(k_b*T)^-1*(3/2)N*k_b*dT for the left side

dS=(3/2)N*T^-1*dT

S=(3/2)N*ln(T) from T_L to T_f, so

S=(3/2)N*ln(T_f/T_L)

and similarly, S=3N*ln(T_f/T_R)

I know that I'm missing a k_b in the end. The entropy on either side of the wall is not equivalent. I'm not sure where I went wrong.