Isothermal Adiabatic Isobaric Isochoric Matching

❓ Question

Match the following lists correctly:

List - I	List - II
(A) Isothermal	(I) ΔW (work done) = 0
(B) Adiabatic	(II) ΔQ (supplied heat) = 0
(C) Isobaric	(III) ΔU (change in internal energy) ≠ 0
(D) Isochoric	(IV) ΔU = 0

---

🖼️ Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Thermodynamic processes
👉 First law of thermodynamics
👉 Internal energy relation.

Main idea:

For ideal gases:

$$\boxed{ \Delta U \propto \Delta T }$$

So internal energy changes only when temperature changes.

---

🔷 Step 1 — Isothermal Process 💯

In isothermal process:

$$T=\text{constant}$$

Hence:

$$\Delta T=0$$

Therefore:

$$\boxed{ \Delta U=0 }$$

So:

$$(A)\rightarrow(IV)$$

---

🔷 Step 2 — Adiabatic Process

In adiabatic process:

$$\boxed{ \Delta Q=0 }$$

No heat exchange occurs.

Thus:

$$(B)\rightarrow(II)$$

---

🔷 Step 3 — Isobaric Process

In isobaric process:

$$P=\text{constant}$$

Temperature generally changes.

Hence:

$$\Delta U\ne0$$

So:

$$(C)\rightarrow(III)$$

---

🔷 Step 4 — Isochoric Process

In isochoric process:

$$V=\text{constant}$$

Work done:

$$W=P\Delta V$$

Since:

$$\Delta V=0$$

Thus:

$$\boxed{ \Delta W=0 }$$

Hence:

$$(D)\rightarrow(I)$$

---

🔷 Step 5 — Final Matching

$$(A)\rightarrow(IV)$$
$$(B)\rightarrow(II)$$
$$(C)\rightarrow(III)$$
$$(D)\rightarrow(I)$$

This matches:

$$\boxed{ \text{Option 4} }$$

---

🔷 Step 6 — JEE Trap Alert 🚨

❌ Isothermal me work done zero maan lena

❌ Isochoric aur adiabatic confuse kar dena

❌ Internal energy ko pressure dependent samajh lena

Remember:

$$\boxed{ \Delta U \text{ depends only on temperature} }$$

---

✅ Final Answer

$$\boxed{ (A)-(IV),\ (B)-(II),\ (C)-(III),\ (D)-(I) }$$

(Option 4)

---

📚 Related Topics

• RC Circuit Phase 90° Condition
• MOI of Combined Bodies Using Parallel Axis
• Find Height Using Refraction in Layered Media