Chemistry Two approaches to calculating entropy differ by factor of two. Why?

AI Thread Summary
The discussion centers on two methods for calculating entropy changes in a thermal system, revealing a discrepancy in results. The first approach yields an entropy change of 0.1484, which is twice the expected value of 0.0740. This error arises from using fixed temperatures in the integrals, neglecting the fact that temperatures T1 and T2 vary during thermalization. The second method correctly accounts for these temperature changes, leading to the accurate calculation of entropy. The conversation concludes with a suggestion to substitute for temperature in the integral to address the initial error.
zenterix
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Homework Statement
A sample of monoatomic ideal gas (##n=1.00\text{mol}##) at ##T_1=300\text{K}## is allowed to thermalize with a second sample of the same ideal gas (##n=1.00\text{mol}##) at ##T_2=350\text{K}##.
Relevant Equations
What is the change in entropy for this process?

Recall that for a monoatomic ideal gas ##C_V=\frac{3}{2}R##.
Here is how I did this problem

Let's call the two samples sample 1 and sample 2.

The change in entropy for sample 1 is

$$\Delta S_1=\int dS_1=\int_{U_1}^{U_1+\Delta U}\frac{1}{T_1}dU\tag{1}$$

$$=\frac{1}{T_1}\Delta U\tag{2}$$

Similarly, ##\Delta S_2=-\frac{1}{T_2}\Delta U##.

Note that I used the fact that ##U## is extensive and conserved so

$$U_1+\Delta U_1+U_2+\Delta U_2=U_1+U_2$$

$$\implies \Delta U_1=-\Delta U_2=\Delta U$$

The entropy change of the system is then

$$\Delta S=\Delta U \left (\frac{1}{T_1}-\frac{1}{T_2}\right )\geq 0\tag{3}$$

$$=\Delta U\left ( \frac{1}{300}-\frac{1}{350}\right )\tag{4}$$

$$\implies \Delta U\geq 0\tag{5}$$

Note that in (1) the integral is defined in such a way that the internal energy of sample 1 is increasing by ##\Delta U## which we've now shown is nonnegative.

Thus, energy (heat in this process) flows from sample 2 to sample 1.

We need to find what ##\Delta U## is.

$$dU=dQ=C_VdT=\frac{3}{2}RdT$$

$$\Delta U=\frac{3}{2}R(T-T_1)=-\frac{3}{2}R(T-T_2)$$

$$\implies T=\frac{T_1+T_2}{2}=325\text{K}$$

where ##T## is the equilibrium temperature of the system.

Thus,

$$\Delta U=\frac{3}{2}R\cdot 25$$

and

$$\Delta S=\Delta U\left (\frac{1}{T_1}-\frac{1}{T_2}\right )$$

$$=\frac{3}{2}R\cdot 25\cdot\left (\frac{1}{300}-\frac{1}{350}\right )$$

$$=0.1484$$

Now, this result seems to be incorrect, and more precisely it seems to be about double the correct result which is 0.0740.

Here is another approach.

The change in entropy for an adiabatic process is

$$\Delta S = \int dS=\int_{T_i}^{T_f} \frac{1}{T} C_VdT$$

$$=C_V\ln{\left (\frac{T_f}{T_i}\right )}$$

$$=\frac{3}{2}R\ln{\left (\frac{T_f}{T_i}\right )}$$

Thus

$$\Delta S_{\text{tot}}=\Delta S_1+\Delta S_2$$

$$=\frac{3}{2}R\left ( \ln{\left ( \frac{325}{300} \right )}+\ln{\left ( \frac{325}{350} \right )} \right )$$

$$=0.0740$$

What is wrong with the first approach?
 
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Temperatures T1 and T2 will change during thermalization. Hence eq. 1 and 2 are wrong as you have used fixed T1 and T2 in the integrals instead of the varying temperatures.
 
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Got it. Is there a way to substitute for ##\frac{1}{T}## in (1) instead of subbing in for ##dU##?
 
As U= C_vT, T=U/C_V.
 
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