Thermal Physics Introduction- solving for entropy and temperature change

In summary: Okay, I'm not sure what you did there, but I don't think it's correct. Remember that the entropy changes have to have opposite signs because the left side is hotter than the right side. So the entropy change on the left is positive, and on the right is negative. So the sum of the two changes is positive, and that's the total change in entropy. It's going in the right direction as you computed, but it's too small. It should be more than 1 J/K. Try again.You're right, I forgot the negative sign. So I got -.370N*k_b (k_b is Boltzmann's constant) for the left and .740N
  • #1
bbolddaslove
8
0

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.
 
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  • #2
bbolddaslove said:

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)\\

Well, you know that T_f - T-L has to be positive and T_f - T_R has to be negative, right? So how can those two differences have the same sign?
You need to get the right T_f first.

so T_f=2T_R-T_L = 1000K-200K = 800K
I'm not sure if i did this right or not
Well, that would be a neat trick; you're starting with 200K and 500K and end up with 800K without doing any work or adding heat to the system?
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.
[/b]

Your S is the change in entropy, not the entropy. So there's a change in S on the left = ΔS_L and a change in the right ΔS_R, and the total entropy change for your system is ΔS_total = ΔS_L + ΔS_R.

So now recompute those entropy changes and then sum. Just don't get a negative result for the net change in S, or Clausius et al will roll over in their graves! :biggrin:
 
  • #3
Okay so I recomputed the temperature using -(Tf-Tr) and got 400. From there i got left S= (3/2)N*ln(2) (Tf/Tl = 2) which gave me 1.04N J/K. On the right I got S=3N*ln(4/5) which was -.669N J/K. I summed them, but I think I'm still missing a k_b somewhere. I crossed them out in the same step i.e.
dS=(k_b*T)^-1*(3/2)N*k_b*dT for the left side
dS=(3/2)N*T^-1*dT

but the answer when multiplied by N without a k_b is too large, isn't it?
 
  • #4
bbolddaslove said:
Okay so I recomputed the temperature using -(Tf-Tr) and got 400. From there i got left S= (3/2)N*ln(2) (Tf/Tl = 2) which gave me 1.04N J/K. On the right I got S=3N*ln(4/5) which was -.669N J/K. I summed them, but I think I'm still missing a k_b somewhere. I crossed them out in the same step i.e.
dS=(k_b*T)^-1*(3/2)N*k_b*dT for the left side
dS=(3/2)N*T^-1*dT

but the answer when multiplied by N without a k_b is too large, isn't it?

I'm not sure what the b is in your equations, but somewhere you dropped something all right.You have the right expressions for the two entropy changes now, except for a constant coefficient. If you just keep track of your coefficients (the N, k, b etc.) you will get the right answer.

Hint: one form of that coefficient is: divide N by some (big) number, then multiply by another number involving R, the universal gas constant. It doesn't involve k or b in that form.

I have my answer waiting if you want to compare w/ yours.
 
  • #5
I got .370N*k_b where k_b is the Boltzmann constant 1.38E-23, does that look like yours?
 
  • #6
Must be absolutely correct! We agree!
 
  • #7
Thank you!
 

1. What is thermal physics?

Thermal physics is a branch of physics that studies the behavior of systems in relation to temperature and heat. It involves the study of how energy is transferred and transformed between different forms, and how this affects the properties of matter.

2. What is entropy?

Entropy is a measure of the disorder or randomness of a system. In thermal physics, it is often described as the amount of energy that is unavailable for doing work. It is closely related to the concept of temperature, as systems tend to become more disordered (higher entropy) as their temperature increases.

3. How do you solve for entropy and temperature change?

The change in entropy (ΔS) can be calculated by dividing the amount of energy transferred (ΔQ) by the temperature (T) at which the transfer occurs: ΔS = ΔQ/T. This equation is known as the Second Law of Thermodynamics. To solve for temperature change, rearrange the equation to T = ΔQ/ΔS.

4. What is the relationship between entropy and temperature?

The relationship between entropy and temperature is described by the Second Law of Thermodynamics, which states that the total entropy of a closed system will always increase over time. This means that as the temperature of a system increases, its entropy will also increase.

5. How is thermal physics important in everyday life?

Thermal physics plays a crucial role in many aspects of our daily lives. It helps us understand how heat is transferred and how different materials respond to changes in temperature. This knowledge is essential in fields such as engineering, materials science, and climate science. It also has practical applications in areas like heating and cooling systems, refrigeration, and energy production.

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