- #1
zenterix
- 774
- 84
- 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?
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?