Give step-by-step solution with explanation and final answer:20-23. In this problem, we will show that Q = c(N) f (E) V" for an ideal gas (Example 20-3).
In Problem 18-42 we showed that the number of

Question

Give step-by-step solution with explanation and final answer:20-23. In this problem, we will show that Q = c(N) f (E) V" for an ideal gas (Example 20-3).
In Problem 18-42 we showed that the number of translational energy states between ¢ and
€ + Ac for a particle in a box can be calculated by considering asphereinn_, n , n_space,

8ma’e
Bp 2 spe id 2
mtn tn =—m—=R
Show that for an N-particle system, the analogous expression is
N 2
8ma”E
Dk nl nl) = ws LS
j=1
or, in more convenient notation
3N 2
8ma’E
2 2
dom=—p—=R
j=l
Thus, instead of dealing with a three-dimensional sphere as we did in Problem 18-42, here
we must deal with a 3N-dimensional sphere. Whatever the formula for the volume of a
3N-dimensional sphere is (it is known), we can at least say that it is proportional to R*".
Show that this proportionality leads to the following expression for ®(E), the number of
states with energy < E,
sma E\
B(E) (25) = c(N)ENRYN
where ¢(N) is a constant whose value depends upon N and V = a*. Now, following the
argument developed in Problem 18-42, show that the number of states between E and
E + AE (which is essentially 2) is given by
Q=c(N)f(E)V¥AE
where f(E) = E¥

VIPSolutions · Accepted Answer

Let's break down the problem and show the derivation step-by-step:

---

### **Step 1: Generalizing to an N-particle System**

For a single particle, the energy states can be represented as a sphere in $ n_x, n_y, n_z $ space:
$$
n_x^2 + n_y^2 + n_z^2 = \frac{8ma^2 \epsilon}{h^2} = R^2
$$

For **N particles**, sum over all particles:
$$
\sum_{j=1}^N (n_{xj}^2 + n_{yj}^2 + n_{zj}^2) = \frac{8ma^2 E}{h^2} = R^2
$$
Or, combining all 3N indices:
$$
\sum_{j=1}^{3N} n_j^2 = \frac{8ma^2 E}{h^2} = R^2
$$

---

### **Step 2: Number of States ≤ E — Volume of a 3N-dimensional Sphere**

The number of states with energy ≤ E is proportional to the **volume of a 3N-dimensional sphere** of radius $ R $:

$$
\Phi(E) \propto R^{3N}
$$

Since 
$$
R^2 = \frac{8ma^2 E}{h^2} \implies R = \left( \frac{8ma^2 E}{h^2} \right)^{1/2}
$$

So,
$$
R^{3N} = \left( \frac{8ma^2 E}{h^2} \right)^{3N/2}
$$

Thus,
$$
\Phi(E) \propto \left( \frac{8ma^2 E}{h^2} \right)^{3N/2}
$$

Or, as given:
$$
\Phi(E) = c(N) E^{3N/2} V^N
$$
where $ c(N) $ is a constant and $ V = a^3 $.

---

### **Step 3: Number of States Between $ E $ and $ E + \Delta E $ ($ \Omega $)**

The number of states in the energy shell $ E $ to $ E + \Delta E $ is:

$$
\Omega = \frac{d\Phi(E)}{dE} \Delta E
$$

From above,
$$
\Phi(E) = c(N) V^N E^{3N/2}
$$
So,
$$
\frac{d\Phi(E)}{dE} = c(N) V^N \cdot \frac{3N}{2} E^{3N/2 - 1}
$$

Set
$$
f(E) = E^{3N/2 - 1}
$$

Therefore,
$$
\Omega = c(N) f(E) V^N \Delta E
$$

---

### **Step 4: Final Answer**

$$
\boxed{
\Omega = c(N) f(E) V^N \Delta E
}
$$
where 
$$
f(E) = E^{3N/2 - 1}
$$

---

### **Summary of Steps:**
1. **Generalize to N particles:** The number of quantum states is determined by the volume of a sphere in 3N-dimensional space.
2. **Proportional to $ R^{3N} $:** Relate energy to the radius, and thus to the number of states.
3. **Differentiate to get states in $ E $ to $ E + \Delta E $:** $ \Omega = \frac{d\Phi}{dE} \Delta E $.
4. **Arrive at final form:** $ \Omega = c(N) f(E) V^N \Delta E $, with $ f(E) = E^{3N/2 - 1} $.

Let me know if you'd like the explicit form for $ c(N) $ or a deeper explanation of any step!

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