A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglec

Q: Why does the pressure decrease in the direction of acceleration?

Due to inertia, the gas molecules tend to resist the forward acceleration and crowd towards the rear of the compartment. This accumulation of molecules at the back creates a higher pressure zone at the rear and a lower pressure zone at the front.

Q: What would happen if gravity was not neglected?

If gravity $g$ were not neglected, there would be a vertical pressure gradient as well, given by $\frac{dP}{dy} = -\rho g$. The total pressure variation would be a combination of both horizontal and vertical gradients: $dP = -\rho a dx - \rho g dy$.

Q: Does this result apply to liquids as well as gases?

Yes, the derivation is completely general for any fluid (liquid or gas) of density $\rho$. In an accelerating container of liquid, the free surface tilts to remain perpendicular to the effective gravity vector $\vec{g}_{\text{eff}} = \vec{g} - \vec{a}$.

Q: Why is the pressure force on the right face written as $(P + dP)A$ instead of just $PA$?

Because the pressure varies continuously with position. If the pressure at position $x$ is $P$, then at a nearby position $x + dx$, the pressure must be $P + dP$, where $dP$ represents the infinitesimal change in pressure over the distance $dx$.

Q: Is the density of the gas constant throughout the compartment?

For an ideal gas, density is proportional to pressure ($P = \rho R T$). However, for moderate accelerations and small compartments, the variation in density is extremely small, and assuming a constant average density $\rho$ yields an excellent approximation that correctly predicts the sign of the pressure gradient.

Question

A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then, the pressure in the compartment is

Options:

(A) same everywhere

(B) lower in front side

(C) lower in rear side

(D) lower in upper side

Sigma Percentile · Accepted Answer

### Introduction: Fluids in Motion and Inertia

Have you ever wondered what happens to the air inside a car when you suddenly hit the gas pedal?

You feel pressed back into your seat, but what about the invisible air molecules surrounding you?

This classic JEE problem invites us to explore the fascinating intersection of fluid mechanics and Newtonian dynamics.

By analyzing a closed compartment containing gas accelerating horizontally, we can uncover how pressure gradients are established in fluids under acceleration.

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### Analyzing the Setup

Let us visualize a closed rectangular compartment filled with a gas of uniform density $ho$.

The compartment is accelerating horizontally to the right with a constant acceleration $a$.

We are instructed to neglect the effect of gravity, which simplifies our analysis to a single dimension along the horizontal $x$-axis.

To find how pressure varies within this compartment, we must zoom in and examine a tiny, representative slice of the gas.

Let us isolate a small vertical element of gas of thickness $dx$ and cross-sectional area $A$ at a distance $x$ from the rear wall.

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### The Master Equation: Applying Newton's Second Law

Every physical element of the gas must accelerate at the same rate $a$ as the compartment itself.

Let us identify the forces acting on our isolated gas element of mass $dm$:

1. **Mass of the element ($dm$):** 
   $$dm = ho \cdot dV = ho \cdot A \cdot dx$$

2. **Pressure Force from the Left ($F_1$):** 
   The gas to the left of our element exerts a forward-pushing force:
   $$F_1 = P \cdot A$$

3. **Pressure Force from the Right ($F_2$):** 
   The gas to the right exerts a backward-pushing force:
   $$F_2 = (P + dP) \cdot A$$

Applying Newton's second law in the horizontal direction:

$$F_{	ext{net}} = F_1 - F_2 = dm \cdot a$$

Substituting our expressions for the forces and mass:

$$P \cdot A - (P + dP) \cdot A = (ho \cdot A \cdot dx) \cdot a$$

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### Simplifying and Solving

Let us expand and simplify the force equation:

$$-dP \cdot A = ho \cdot A \cdot dx \cdot a$$

Dividing both sides by the common factor $A$:

$$-dP = ho \cdot a \cdot dx$$

Rearranging this gives us the pressure gradient along the direction of acceleration:

$$\frac{dP}{dx} = -ho \cdot a$$

Since density $ho$ and acceleration $a$ are inherently positive physical quantities, the product $ho a$ is positive.

Therefore, the derivative is negative:

$$\frac{dP}{dx} < 0$$

This mathematical result tells us that as we move in the positive $x$-direction (from the rear of the compartment to the front), the pressure $P$ must decrease.

Consequently, the pressure is **lower at the front side** and higher at the rear side.

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### The Intuitive Picture: Pseudo-Forces

If you analyze this from the non-inertial reference frame of the accelerating compartment, a pseudo-force acts on every gas molecule.

This pseudo-force is directed opposite to the acceleration, i.e., towards the rear wall:

$$F_{	ext{pseudo}} = -m \cdot a$$

This causes the gas molecules to crowd towards the back of the compartment, creating a region of higher density and pressure at the rear.

To prevent all the gas from collapsing to the back, a pressure gradient must develop to balance this pseudo-force, resulting in lower pressure at the front.

Thus, the correct option is **(b)**.

Animated Solution for Physics - Properties of Solids and Liquids: A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then, the pressure in the compartment is

Select Answer:

Visualized Solution

Visualizing the Accelerating Compartment

Selecting a Fluid Element

Identifying the Mass of the Element

Identifying Pressure Forces

Applying Newton's Second Law

Simplifying the Force Equation

Interpreting the Pressure Gradient

Real-World Analogy and Gravity Effects

The Sigma Insight: Fluid Pressure and Pascal's Law

Introduction

Analyzing the Setup

The Master Equation

Simplifying and Solving

The Intuitive Picture

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