- #1
runinfang
- 4
- 0
- Homework Statement
- One mole of a monatomic ideal gas and one mole of an ideal van der Waals fluid with c = 3/2 are contained separately in vessels of fixed volumes v1 and v2. The temperature of the ideal gas is T1 and that of the van der Waals fluid is T2. It is desired to bring the ideal gas to temperature T2, maintaining the total energy constant. What is the final temperature of the van der Waals fluid? What restrictions apply among the parameters (T 1, T2 , a, b, v1, v2) if it is to be possible to design an engine to accomplish this temperature inversion (assuming, as always, that no external system is to be altered in the process)?
- Relevant Equations
- Internal energy of an ideal gas: ## u = CRT##
Internal energy of the van der Waals fluid: ##u = CRT - \frac{a}{v}##
Since the energy variation is zero:
$$
\Delta U = \Delta U_{1} + \Delta U_{2} = 0
$$
The energy for a monatomic ideal gas is ## u = CRT##, and the energy for a Van der Waals gas is
$$
u = CRT - \frac{a}{v},
$$
obtained through
$$
\frac{1}{T} = \frac{CR}{a + \frac{a}{v}}.
$$
Summing the internal energies:
$$
\text{CRT}_{1} + \text{CRT}_{2} - \frac{a}{v_{2}} = \text{CRT}_{1} + \text{CRT}_{f} - \frac{a}{v_{2}}
$$
$$
T_{1} + T_{2} = T_{2} + T_{f}
$$
$$
T_{1} = T_{f}
$$
This is the first part of the problem. I'm not sure about the answer... I’ve seen other solutions that gave quite different results, but I don’t know how to arrive at those or if they are correct.
To identify the second question, to identify the constraints, ##\Delta S \geq 0 \##. I tried using molar entropy to simplify.
The change in molar entropy for an ideal gas is:
$$
\Delta S = cR \ln\left(\frac{T}{T_{0}}\right) + R \ln\left(\frac{v}{v_{0}}\right)
$$
Applying it to the problem, we have:
$$
\Delta S = cR \ln\left(\frac{T_{2}}{T_{1}}\right) + R \ln\left(\frac{v_{1}}{v_{1}}\right)
$$
Since the volume does not vary, we have:
$$
\Delta S = cR \ln\left(\frac{T_{2}}{T_{1}}\right)
$$
The change in entropy for a Van der Waals gas is:
$$
\Delta S = R \ln\left(\frac{v_{1} - b}{v_{0} - b}\right) + cR \ln\left(\frac{T_{1}}{T_{0}}\right)
$$
Applying it to the problem, we have:
$$
\Delta S = R \ln\left(\frac{v_{2} - b}{v_{2} - b}\right) + cR \ln\left(\frac{T_{1}}{T_{2}}\right)
$$
$$
\Delta S = cR \ln\left(\frac{T_{1}}{T_{2}}\right)
$$
But if I sum the entropies, it only results in zero, and I can't analyze the constraints.
$$
\Delta U = \Delta U_{1} + \Delta U_{2} = 0
$$
The energy for a monatomic ideal gas is ## u = CRT##, and the energy for a Van der Waals gas is
$$
u = CRT - \frac{a}{v},
$$
obtained through
$$
\frac{1}{T} = \frac{CR}{a + \frac{a}{v}}.
$$
Summing the internal energies:
$$
\text{CRT}_{1} + \text{CRT}_{2} - \frac{a}{v_{2}} = \text{CRT}_{1} + \text{CRT}_{f} - \frac{a}{v_{2}}
$$
$$
T_{1} + T_{2} = T_{2} + T_{f}
$$
$$
T_{1} = T_{f}
$$
This is the first part of the problem. I'm not sure about the answer... I’ve seen other solutions that gave quite different results, but I don’t know how to arrive at those or if they are correct.
To identify the second question, to identify the constraints, ##\Delta S \geq 0 \##. I tried using molar entropy to simplify.
The change in molar entropy for an ideal gas is:
$$
\Delta S = cR \ln\left(\frac{T}{T_{0}}\right) + R \ln\left(\frac{v}{v_{0}}\right)
$$
Applying it to the problem, we have:
$$
\Delta S = cR \ln\left(\frac{T_{2}}{T_{1}}\right) + R \ln\left(\frac{v_{1}}{v_{1}}\right)
$$
Since the volume does not vary, we have:
$$
\Delta S = cR \ln\left(\frac{T_{2}}{T_{1}}\right)
$$
The change in entropy for a Van der Waals gas is:
$$
\Delta S = R \ln\left(\frac{v_{1} - b}{v_{0} - b}\right) + cR \ln\left(\frac{T_{1}}{T_{0}}\right)
$$
Applying it to the problem, we have:
$$
\Delta S = R \ln\left(\frac{v_{2} - b}{v_{2} - b}\right) + cR \ln\left(\frac{T_{1}}{T_{2}}\right)
$$
$$
\Delta S = cR \ln\left(\frac{T_{1}}{T_{2}}\right)
$$
But if I sum the entropies, it only results in zero, and I can't analyze the constraints.
Last edited:
