Chemistry How to integrate a sum of two entropy differentials?

  • Thread starter Thread starter zenterix
  • Start date Start date
  • Tags Tags
    Entropy
AI Thread Summary
The discussion focuses on integrating the sum of two entropy differentials in a thermodynamic context. The initial equations demonstrate the relationship between heat transfer and entropy change, emphasizing the need for reversible processes in calculations. Participants clarify that the temperatures T1 and T2 are not constant and that the integration limits should reflect the varying temperatures of the blocks. The conversation concludes that while the integration approach is valid, it ultimately leads to the same result as the initial solution, reinforcing the concept that the total entropy change is positive, indicating irreversibility. The complexities of these thermodynamic concepts highlight the challenges in understanding heat exchange and entropy.
zenterix
Messages
774
Reaction score
84
Homework Statement
Two blocks of the same metal are of the same size but are at different temperatures ##T_1## and ##T_2##. These blocks of metal are brought together and allowed to come to the same temperature.
Relevant Equations
Show that the entropy change is given by

$$\Delta S=C_P\ln{\left [ \frac{(T_1+T_2)^2}{4T_1T_2} \right ]}$$

if ##C_P## is constant.
Here is a way to solve the problem.

Since ##dq_1=-dq_2## then

$$\int_{T_1}^T C_PdT=-\int_{T_2}^T C_PdT\tag{1}$$

$$\implies T=\frac{T_1+T_2}{2}\tag{2}$$

$$dq_1=C_PdT\tag{3}$$

$$dS_1=\frac{dq_1}{T}=\frac{C_P}{T}dT\tag{4}$$

$$\Delta S_1=\int_{T_1}^T\frac{C_P}{T}dT=C_P\ln{\frac{T}{T_1}}\tag{5}$$

$$dq_2=C_PdT\tag{6}$$

$$dS_2=\frac{dq_2}{T}=\frac{C_P}{T}dT\tag{7}$$

$$\Delta S_1=\int_{T_2}^T\frac{C_P}{T}dT=C_P\ln{\frac{T}{T_2}}\tag{8}$$

$$\Delta S_{sys}=\Delta S_1+\Delta S_2=C_P\ln{\frac{T^2}{T_1T_2}}=C_P\ln{\frac{(T_1+T_2)^2}{4T_1T_2}}\tag{9}$$

My question is about solving it a different way, if possible.

Can we write

$$dS_{sys}=dS_1+dS_2=\frac{dq_1}{T_1}+\frac{dq_2}{T_2}\tag{10}$$

$$=\frac{dq_1}{T_1}-\frac{dq_1}{T_2}\tag{11}$$

is this expression correct?

I am really not sure. If it is, I am not sure what it means to integrate the right-hand side. In particular, what would the limits of integration be?
 
Physics news on Phys.org
zenterix said:
Can we write

$$dS_{sys}=dS_1+dS_2=\frac{dq_1}{T_1}+\frac{dq_2}{T_2}\tag{10}$$

$$=\frac{dq_1}{T_1}-\frac{dq_1}{T_2}\tag{11}$$

is this expression correct?

I am really not sure. If it is, I am not sure what it means to integrate the right-hand side. In particular, what would the limits of integration be?
The 2 blocks don't have constant temperatures.
T1 and T2 are only their starting temperatures.
Try again. Then you should see what the limits of integration have to be.
 
Philip Koeck said:
The 2 blocks don't have constant temperatures.
T1 and T2 are only their starting temperatures.
Try again. Then you should see what the limits of integration have to be.
I am aware that the temperatures aren't constant.

Indeed, the expression ##\frac{dq_1}{T_1}-\frac{dq_1}{T_2}## isn't correct because I defined ##T_1## and ##T_2## as constant initial temperatures.

In the correct solution I showed, when I write ##\int_{T_1}^T\frac{C_P}{T}dT##, the ##T##, as far as I understand, is actually the external temperature. That is, it is the temperature of the external "reservoir" for a reversible process.

Am I correct to say that implicit in this problem is the assumption that the heat exchange happens reversibly?

I am really not sure. The fact that the calculated entropy change is positive means this entire process can occur spontaneously.

After all, the differential ##dS=\frac{dq}{T}## is only true for reversible processes.

If I write ##\frac{dq_1}{t_1}-\frac{dq_1}{t_2}## then ##t_1## and ##t_2## denote external temperatures. Would this be correct?

I find it incredible how difficult these concepts are even though they look so simple.
 
zenterix said:
I am aware that the temperatures aren't constant.

Indeed, the expression ##\frac{dq_1}{T_1}-\frac{dq_1}{T_2}## isn't correct because I defined ##T_1## and ##T_2## as constant initial temperatures.

In the correct solution I showed, when I write ##\int_{T_1}^T\frac{C_P}{T}dT##, the ##T##, as far as I understand, is actually the external temperature. That is, it is the temperature of the external "reservoir" for a reversible process.

Am I correct to say that implicit in this problem is the assumption that the heat exchange happens reversibly?

I am really not sure. The fact that the calculated entropy change is positive means this entire process can occur spontaneously.

After all, the differential ##dS=\frac{dq}{T}## is only true for reversible processes.

If I write ##\frac{dq_1}{t_1}-\frac{dq_1}{t_2}## then ##t_1## and ##t_2## denote external temperatures. Would this be correct?

I find it incredible how difficult these concepts are even though they look so simple.
Yes, you do have to consider a reversible process to calculate ΔS.
You could imagine an environment that has all the temperatures between T1 and T2 so that the temperature change can occur reversibly, but that just means that you only need to think of the temperatures of the two blocks anyway.
Your equations 10 and 11 are correct if T1 stands for the varying temperature of block 1 and T2 for the varying temperature of block 2. When you integrate you just use the starting and final temperatures of each block as limits, so you need two separate integrals.
I would say you end up with the same as the first solution you give.
 
zenterix said:
Am I correct to say that implicit in this problem is the assumption that the heat exchange happens reversibly?

I am really not sure. The fact that the calculated entropy change is positive means this entire process can occur spontaneously.

After all, the differential ##dS=\frac{dq}{T}## is only true for reversible processes.

If I write ##\frac{dq_1}{t_1}-\frac{dq_1}{t_2}## then ##t_1## and ##t_2## denote external temperatures. Would this be correct?

I find it incredible how difficult these concepts are even though they look so simple.
I agree it's strange.

You need an (imagined) reversible process to calculate the entropy change for each block, but in the end you find that the total entropy change is positive, showing that the process i irreversible.

This only makes sense because entropy is a state function.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top