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Question

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VIPSolutions · Accepted Answer

Let's solve the questions step by step for an **ideal gas**.

---

## (1) Prove the following equalities for an *ideal gas*.

### (a) $\left(\frac{\partial T}{\partial V}\right)_{p} \left(\frac{\partial V}{\partial p}\right)_{T} \left(\frac{\partial p}{\partial T}\right)_{V} = -1$

#### **Explanation:**
This uses the **cyclic rule** for partial derivatives:
$$
\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1
$$

Let:
- $x = T$
- $y = V$
- $z = p$

Then,
$$
\left(\frac{\partial T}{\partial V}\right)_p \left(\frac{\partial V}{\partial p}\right)_T \left(\frac{\partial p}{\partial T}\right)_V = -1
$$

#### **Final Answer (a):**
$$
\boxed{-1}
$$

---

### (b) $\left(\frac{\partial V}{\partial n}\right)_{p,T} = \frac{1}{\left(\frac{\partial n}{\partial V}\right)_{p,T}}$

#### **Explanation:**
This is a property of **inverse partial derivatives** (reciprocal rule):

If $f(x, y, z) = 0$, then:
$$
\left(\frac{\partial x}{\partial y}\right)_z = \frac{1}{\left(\frac{\partial y}{\partial x}\right)_z}
$$

So,
$$
\left(\frac{\partial V}{\partial n}\right)_{p,T} = \frac{1}{\left(\frac{\partial n}{\partial V}\right)_{p,T}}
$$

#### **Final Answer (b):**
$$
\boxed{\left(\frac{\partial V}{\partial n}\right)_{p,T} = \frac{1}{\left(\frac{\partial n}{\partial V}\right)_{p,T}}}
$$

---

### (c) $C_p - C_v = nR$ if $C_p - C_v = T \left(\frac{\partial p}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_p$

#### **Explanation:**
For an ideal gas,
$$
pV = nRT
$$

Calculate the partial derivatives:
- $\left(\frac{\partial p}{\partial T}\right)_V = \frac{nR}{V}$
- $\left(\frac{\partial V}{\partial T}\right)_p = \frac{nR}{p}$

So,
$$
C_p - C_v = T \left(\frac{\partial p}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_p
$$
$$
= T \cdot \frac{nR}{V} \cdot \frac{nR}{p}
$$

But, from the ideal gas law, $pV = nRT$ so $V = \frac{nRT}{p}$

$$
\left(\frac{\partial p}{\partial T}\right)_V = \frac{nR}{V}
$$
$$
\left(\frac{\partial V}{\partial T}\right)_p = \frac{nR}{p}
$$

Substitute these in:

$$
C_p - C_v = T \cdot \frac{nR}{V} \cdot \frac{nR}{p}
$$

But, for one mole ($n = 1$), and substituting from $pV = nRT$:

$$
\frac{nRT}{p} = V
\implies \frac{1}{V} = \frac{p}{nRT}
$$

So,

$$
T \cdot \frac{nR}{V} \cdot \frac{nR}{p}
= T \cdot nR \cdot \frac{1}{V} \cdot \frac{nR}{p}
= T \cdot \frac{nR}{V} \cdot \frac{nR}{p}
= T \cdot \frac{nR}{V} \cdot \frac{nR}{p}
$$

But let's simplify:

$$
C_p - C_v = T \cdot \frac{nR}{V} \cdot \frac{nR}{p}
$$

But for $n = 1$:

$$
C_p - C_v = T \cdot \frac{R}{V} \cdot \frac{R}{p}
= \frac{TR^2}{Vp}
$$

But $pV = RT \implies Vp = RT$:

$$
C_p - C_v = \frac{TR^2}{RT} = R
$$

For $n$ moles, it's $nR$:

#### **Final Answer (c):**
$$
\boxed{C_p - C_v = nR}
$$

---

## **Summary Table**

| Part | Final Answer |
|------|--------------|
| (a)  | $-1$       |
| (b)  | $\left(\frac{\partial V}{\partial n}\right)_{p,T} = \frac{1}{\left(\frac{\partial n}{\partial V}\right)_{p,T}}$ |
| (c)  | $C_p - C_v = nR$ |

Let me know if you'd like each step shown in even more detail!

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