Match the thermodynamic processes taking place in a system with the correct conditions. In the table

Q: Why is the change in internal energy zero for an isothermal process? Doesn't adding heat increase internal energy?

For an ideal gas, the internal energy depends *exclusively* on its absolute temperature. In an isothermal process, the temperature is forced to remain constant. Therefore, any heat added to the system is entirely converted into work done by the gas, leaving the internal energy unchanged.

Q: In an adiabatic process, if no heat is supplied ($\Delta Q = 0$), how can the gas do work?

The gas does work at the expense of its own internal energy! According to the First Law, $0 = \Delta U + \Delta W$, which means $\Delta W = -\Delta U$. If the gas expands and does positive work, its internal energy must decrease, leading to a drop in temperature.

Q: Why is $\Delta W = 0$ in an isochoric process?

Work done by a gas is defined as the integral of pressure over the change in volume ($W = \int p \, dV$). In an isochoric process, the volume is strictly constant, meaning $dV = 0$. Without any expansion or compression, the gas cannot perform mechanical work on its surroundings.

Q: For the isobaric process, why are all three quantities ($\Delta Q, \Delta U, \Delta W$) non-zero?

In an isobaric process, the pressure is constant, but the volume changes, so work is done ($\Delta W = p\Delta V \neq 0$). Because the volume changes at constant pressure, the temperature must also change (from $pV = nRT$), meaning internal energy changes ($\Delta U \neq 0$). To achieve both work and a temperature change, heat must be exchanged ($\Delta Q \neq 0$).

Q: Is the First Law of Thermodynamics valid for all these processes?

Absolutely! The First Law ($\Delta Q = \Delta U + \Delta W$) is a statement of the conservation of energy. It is a universal law that applies to every single thermodynamic process. The specific conditions of each process just dictate which terms in the equation become zero or how they relate to each other.

Question

Match the thermodynamic processes taking place in a system with the correct conditions. In the table, $\Delta Q$ is heat supplied, $\Delta W$ is work done and $\Delta U$ is change in internal energy of the system.

List-I:

P. Adiabatic

Q. Isothermal

R. Isochoric

S. Isobaric

List-II:

1. $\Delta W = 0$

2. $\Delta Q = 0$

3. $\Delta U 
eq 0, \Delta W 
eq 0$ and $\Delta Q 
eq 0$

4. $\Delta U = 0$

Sigma Percentile · Accepted Answer

### Decoding Thermodynamic Processes

Thermodynamics is the beautiful study of how energy moves and transforms. At the heart of this entire subject lies a single, incredibly powerful equation: **The First Law of Thermodynamics**.

Mathematically, it is written as:

$$\Delta Q = \Delta U + \Delta W$$

This equation is simply the law of conservation of energy applied to a thermodynamic system. It states that any heat ($\Delta Q$) supplied to a system must either go into changing the system's internal energy ($\Delta U$) or be used by the system to do mechanical work ($\Delta W$) on its surroundings. By applying constraints to this master equation, we can derive the four fundamental thermodynamic processes.

### The Adiabatic Process: The Insulated Journey

Imagine a gas trapped inside a perfectly insulated cylinder. The walls are so thick and thermally resistant that absolutely no heat can enter or escape the system. This is the defining condition of an **adiabatic process**.

Because the system is thermally isolated, the heat exchange is strictly zero:

$$\Delta Q = 0$$

If we plug this into the First Law, we get $0 = \Delta U + \Delta W$, which means $\Delta W = -\Delta U$. This tells us a fascinating physical story: if an adiabatically isolated gas expands and does positive work on its surroundings, it must pay for that work using its own internal energy. As a result, its internal energy drops, and the gas cools down dramatically. This is exactly why the air rushing out of a highly pressurized bicycle tire feels cold!

### The Isothermal Process: The Constant Temperature

Now, let's change the setup. Imagine the cylinder is made of perfectly conducting walls and is submerged in a massive thermal bath. If we compress or expand the gas incredibly slowly, the gas will constantly exchange heat with the bath to ensure its temperature never changes.

'Iso' means same, and 'thermal' means temperature. Thus, the defining condition is:

$$T = 	ext{constant} \implies \Delta T = 0$$

For an ideal gas, the internal energy is a function of temperature alone ($U = \frac{f}{2}nRT$). If the temperature doesn't change, the internal energy remains perfectly static. Therefore, the change in internal energy is zero:

$$\Delta U = 0$$

In this scenario, the First Law becomes $\Delta Q = \Delta W$. Any heat added to the system is entirely converted into work done by the gas.

### The Isochoric Process: The Locked Volume

Let's lock the piston in place so it cannot move. The gas is now trapped in a rigid, fixed volume. We can heat it up or cool it down, but it cannot expand or compress.

'Choric' relates to space or volume. The defining condition is:

$$V = 	ext{constant} \implies \Delta V = 0$$

Mechanical work in thermodynamics is defined as the integral of pressure over the change in volume ($W = \int p \, dV$). If the volume cannot change, the gas is physically incapable of doing any mechanical work. Therefore:

$$\Delta W = 0$$

Here, the First Law simplifies to $\Delta Q = \Delta U$. If you heat a gas at constant volume, 100% of that thermal energy goes directly into increasing the internal energy, causing the temperature and pressure to spike rapidly.

### The Isobaric Process: The Constant Pressure

Finally, imagine a vertical cylinder with a freely moving piston that has a constant weight resting on it. The pressure inside the gas is determined entirely by the atmospheric pressure plus the weight of the piston. As long as the weight doesn't change, the pressure remains constant.

'Baric' relates to pressure. The defining condition is:

$$p = 	ext{constant} \implies \Delta p = 0$$

In this process, if we heat the gas, it will expand to maintain the constant pressure. Because it expands, it does work ($\Delta W = p\Delta V 
eq 0$). Because it expands at constant pressure, its temperature must increase according to the ideal gas law ($pV = nRT$), meaning its internal energy changes ($\Delta U 
eq 0$). And to fuel both the expansion work and the temperature rise, heat must be actively supplied ($\Delta Q 
eq 0$).

Therefore, in an isobaric process, **none of the three quantities are zero**.

### The Final Verdict

By understanding the physical constraints of each process, the matching becomes trivial:
*   **Adiabatic (I)** means no heat transfer, matching with **$\Delta Q = 0$ (B)**.
*   **Isothermal (II)** means constant temperature and thus no change in internal energy, matching with **$\Delta U = 0$ (D)**.
*   **Isochoric (III)** means constant volume and thus no work done, matching with **$\Delta W = 0$ (A)**.
*   **Isobaric (IV)** means constant pressure where work is done, temperature changes, and heat is exchanged, matching with **$\Delta U 
eq 0, \Delta W 
eq 0, \Delta Q 
eq 0$ (C)**.

This logical breakdown is the key to mastering thermodynamic cycles and heat engines!

Animated Solution for Physics - Thermodynamics: Match the thermodynamic processes taking place in a system with the correct conditions. In the table, $Δ Q$ is heat supplied, $Δ W$ is work done and $Δ U$ is change in internal energy of the system.

List-I

List-II

Select Matching Pairs:

Visualized Solution

$The First Law$

$Adiabatic Process$

$Isothermal Process$

$Internal Energy in Isothermal$

$Isochoric Process$

$Work Done in Isochoric$

$Isobaric Process$

$Isobaric Variables$

$Final Matching$

The Sigma Insight: Thermodynamic Processes

Decoding Thermodynamic Processes

The Adiabatic Process

The Isothermal Process

The Isochoric Process

The Isobaric Process

The Final Verdict

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