Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: $1\ Pa = 1\ N\ m^{-2}$. If mass of air at sea level is $1034\ g\ cm^{-2}$, calculate the pressure in pascal.

Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown belo

Exercises · 1.13

1\ Pa = 1\ N\ m^{-2}

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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The key idea is to convert the given mass per area into force per area by multiplying by $g$ , then convert all units to SI (kg, m, s) to get pressure in pascals. The final pressure is approximately $1.01 \times 10^{5}\ \text{Pa}$ .

Why this works

Pressure is defined as force per unit area. The given value — $1034\ \text{g cm}^{-2}$ — is a mass per area, not a force per area. To turn it into pressure, we need to multiply by the acceleration due to gravity ( $g = 9.8\ \text{m s}^{-2}$ ), because weight (force) = mass × $g$ .

So the pressure at sea level is simply the weight of the column of air above each square centimetre. The problem gives us that mass per area directly. Our job is to convert that into pascals, the SI unit of pressure.

Watch out

A common mistake is to forget that $1034\ \text{g cm}^{-2}$ is a mass density per area, not a force. You must multiply by $g$ to get pressure. Also, don't forget to convert grams to kilograms and centimetres to metres — the pascal uses SI base units.

Step-by-step solution

1. Write down what we know

We are given:

Mass per area: $m/A = 1034\ \text{g cm}^{-2}$
Acceleration due to gravity: $g = 9.8\ \text{m s}^{-2}$ (standard value)
$1\ \text{Pa} = 1\ \text{N m}^{-2} = 1\ \text{kg m}^{-1}\text{s}^{-2}$

Pressure $P = \frac{F}{A} = \frac{mg}{A} = \left(\frac{m}{A}\right) \times g$

2. Convert mass from grams to kilograms

$1\ \text{kg} = 1000\ \text{g}$ , so:

1034\ \text{g} = \frac{1034}{1000}\ \text{kg} = 1.034\ \text{kg}

3. Convert area from cm² to m²

$1\ \text{m} = 100\ \text{cm}$ , so $1\ \text{m}^2 = (100)^2\ \text{cm}^2 = 10^4\ \text{cm}^2$

Therefore:

1\ \text{cm}^2 = 10^{-4}\ \text{m}^2

So the mass per area in SI units becomes: …

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🧠 The Core Concept

Pressure is force per unit area.

The given value ( $1034\ g\ cm^{-2}$ ) is mass per unit area, not force per unit area.

To get pressure in pascals, you must convert mass to force (using $F = mg$ ) and area to SI units ( $m^2$ ).

✗ Common Mistake #1: Treating mass as force directly

What students do:

They take $1034\ g\ cm^{-2}$ and try to directly convert grams to kg and cm² to m², then call it pressure.

Why it’s wrong:

Pressure requires force (Newtons), not mass.

$1\ N = 1\ kg\ m\ s^{-2}$ . You must multiply mass by $g = 9.8\ m\ s^{-2}$ (or $10\ m\ s^{-2}$ if the problem allows approximation).

✓ How to avoid:

Always ask: “Is this mass or force?”

If it’s mass, multiply by $g$ to get weight (force).

✗ Common Mistake #2: Forgetting to convert $g$ to $kg$

What students do:

They keep $1034\ g$ as $1034$ and only convert area.

Why it’s wrong:

SI unit of mass is kg, not g.

$1\ g = 10^{-3}\ kg$ .

✓ How to avoid:

Write every conversion factor explicitly:

$1034\ g = 1034 \times 10^{-3}\ kg = 1.034\ kg$

✗ Common Mistake #3: Forgetting to convert $cm^2$ to $m^2$

What students do:

They convert $1\ cm^2 = 10^{-2}\ m^2$ (wrong!) or just ignore area conversion.

Why it’s wrong:

$1\ cm = 10^{-2}\ m$ , so

$1\ cm^2 = (10^{-2})^2\ m^2 = 10^{-4}\ m^2$

✓ How to avoid:

Remember: area conversion factor = (length conversion factor)²

Always square the linear conversion.

✗ Common Mistake #4: Mixing up $g\ cm^{-2}$ with $N\ m^{-2}$

What students do:

They think $g\ cm^{-2}$ is already a pressure unit.

Why it’s wrong:

$g\ cm^{-2}$ is mass per area, not force per area.

Pressure in SI is $N\ m^{-2}$ (pascal).

✓ How to avoid:

Check the units:

Mass/area → need × $g$
Force/area → already pressure

✓ Correct Step-by-Step Solution

Step 1: Write given value …

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←Previous question1.12 — If the density of methanol is 0.793kgL⁻¹, what is its volume needed fo…→Next question1.14 — What is the SI unit of mass? How is it defined?

Q.Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: $1\ Pa = 1\ N\ m^{-2}$ . If mass of air at sea level is $1034\ g\ cm^{-2}$ , calculate the pressure in pascal.

Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

The Intuition: Same Quantity, Different Labels

The Precise Statement

How to Convert: The Only Rule You Need

The Reverse: Metres to Centimetres

Common Conversion Factors You'll Use

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

1. The Core Idea: Physical Quantities Have Dimensions

2. The Principle of Dimensional Homogeneity

Why must this hold?

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

Why does this work?

4. Why We Can Convert Units Using Dimensional Analysis

Why this works: …

Why this works

Step-by-step solution

Method: Dimensional Analysis (Factor-Label Method)

Step 1: Understand what is given

Step 2: Write the starting quantity

Step 3: Apply conversion factors

Step 4: Set up the chain

Step 5: Cancel units and calculate …

🧠 The Core Concept

✗ Common Mistake #1: Treating mass as force directly

✗ Common Mistake #2: Forgetting to convert $g$ to $kg$

✗ Common Mistake #3: Forgetting to convert $cm^2$ to $m^2$

✗ Common Mistake #4: Mixing up $g\ cm^{-2}$ with $N\ m^{-2}$

✓ Correct Step-by-Step Solution

Q.Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1 Pa=1 N m−21\ Pa = 1\ N\ m^{-2}1 Pa=1 N m−2. If mass of air at sea level is 1034 g cm−21034\ g\ cm^{-2}1034 g cm−2, calculate the pressure in pascal.

Why this works

Step-by-step solution

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Q.Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: $1\ Pa = 1\ N\ m^{-2}$ . If mass of air at sea level is $1034\ g\ cm^{-2}$ , calculate the pressure in pascal.

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