Chemistry Bridge Course - Lecture 2 | CHEMCA JEE & NEET

CHEMCA

JEE & NEET Academy by Abhishek Sengar

Lecture 2 Physical Foundations Target: Master Unit Conversion & Basic Logarithms

Important Units, Conversions & Logarithms

Welcome to Lecture 2 of the CHEMCA Bridge Course! In this session, Abhishek Sengar Sir breaks down the 7 Fundamental SI Units, essential derived physical constants (Mass, Volume, Pressure, Energy, Temperature), and the vital mathematical foundation of Logarithms ($\log_{10}$ and $\ln$) used in Chemistry.

Video Lecture Broadcast

Instructor: Abhishek Sengar Sir Published: April 14, 2026 Subject: Units & Math Tools

Interactive Lecture Timestamps

Click any topic to skip the video directly to that specific concept explanation.

In-Depth Lecture Notes & Summary

Significance of Units & Standards

In Physical Chemistry, numeric quantities are meaningless without their corresponding units. Units ensure Standardization across the globe, avoidance of catastrophic engineering calculation errors, and form the absolute core foundation of Stoichiometry and Gas Laws.

The SI System (7 Fundamental Units)

The International System of Units (SI) adopts 7 Fundamental Physical Quantities. All other physical units (such as density, area, volume, velocity) are mathematically derived from these:

Physical Quantity	Fundamental SI Unit	Symbol
Mass	Kilogram	kg
Length	Meter	m
Time	Second	s
Temperature	Kelvin	K
Amount of Substance	Mole	mol
Electric Current	Ampere	A
Luminous Intensity	Candela	cd

Core Quantities & Their Conversions

1. Mass Equivalents

For atomic and subatomic states, we use the Atomic Mass Unit (amu), represented simply as $u$:

• $1 \text{ kg} = 10^3 \text{ g} = 10^6 \text{ mg} = 10^9 \text{ \mu g}$

• $1 \text{ metric ton} = 1000 \text{ kg}$

• $1 \text{ u (or amu)} = \frac{1}{12} \text{ mass of one Carbon-12 atom} \approx 1.66 \times 10^{-24} \text{ g}$

2. Volume Equivalents

Note that cubic centimeter ($cm^3$) is also called cc (common in fuel engine calculations) and is identical to $mL$:

• $1 \text{ m}^3 = 1000 \text{ L} = 10^6 \text{ mL}$

• $1 \text{ L} = 1 \text{ dm}^3 = 1000 \text{ mL} = 1000 \text{ cm}^3\text{ (cc)}$

3. Pressure Equivalents

Pressure Conversions are vital in Gas Laws ($PV = nRT$):

• $1 \text{ atm} = 101325 \text{ Pa} = 1.01325 \text{ bar}$

• $1 \text{ atm} = 760 \text{ mmHg} = 760 \text{ Torr}$

• $1 \text{ bar} = 10^5 \text{ Pa} = 100 \text{ kPa}$

Problem from Video: Convert $380 \text{ mmHg}$ to $atm$.
$$\text{Value in atm} = \frac{380 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.5 \text{ atm}$$

4. Energy Equivalents

Used in Thermodynamics & Atomic Structure ($1 \text{ eV}$ is the work done moving a single electron across $1 \text{ V}$ potential):

• $1 \text{ cal} \approx 4.184 \text{ J} \approx 4.2 \text{ J}$

• $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$

• $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$

5. Temperature Scales ($^{\circ}C$, $K$, $^{\circ}F$)

Kelvin is the absolute thermodynamic scale. Absolute zero temperature represents $0 \text{ K} = -273.15^{\circ}C$. We do not write Degree ($^{\circ}$) for Kelvin because it is an absolute scale.

$$\text{Kelvin (K)} = T(^{\circ}\text{C}) + 273.15$$

$$\text{Fahrenheit (}^{\circ}\text{F)} = \frac{9}{5} \times T(^{\circ}\text{C}) + 32 = 1.8 \times T(^{\circ}\text{C}) + 32$$

Video Challenge Question: Find absolute zero temperature ($0 \text{ K}$ or $-273.15^\circ C$) in Fahrenheit ($^{\circ}F$):
$$T(^{\circ}F) = 1.8 \times (-273.15) + 32 = -491.67 + 32 = -459.67^{\circ}F$$

Understanding Logarithms (Log & Natural Log)

Logarithms are heavily used in thermodynamics, chemical kinetics, chemical equilibrium, and pH calculations. We distinguish between base-10 logarithms ($\log$) and natural base-e logarithms ($\ln$):

$$\ln(x) = \log_e(x) = 2.303 \log_{10}(x) = 2.303 \log(x)$$

Essential Logarithmic Identities

$$\log(a \times b) = \log a + \log b$$

$$\log\left(\frac{a}{b}\right) = \log a - \log b$$

$$\log(a^b) = b \log a$$

Log Values to Memorize (Base-10)

Memorizing these 4 primary prime values ($\log 2, \log 3, \log 5, \log 7$) allows you to derive almost any other log value from 1 to 10:

$\log 1$ 0.000

$\log 2$ (Prime) 0.301

$\log 3$ (Prime) 0.477

$\log 4$ ($2\log 2$) 0.602

$\log 5$ (Prime) 0.699

$\log 6$ ($\log 2+\log 3$) 0.778

$\log 7$ (Prime) 0.845

$\log 8$ ($3\log 2$) 0.903

$\log 9$ ($2\log 3$) 0.954

$\log 10$ 1.000

What is an Antilog?

Antilogarithm is simply the inverse of logarithmic operation. $$\text{If } \log_{10}(x) = y \implies x = \text{antilog}(y) = 10^y$$

CHEMCA Unit Converter

Instantly perform the physical conversions discussed in Lecture 2.

Physical Quantity

Input Value

From

Converted Value 1.000

Interactive Log Helper

Explore log math and identities live.

Input Number ($x$)

Base 10 ($\log x$) 0.7782

Natural Log ($\ln x$) 1.7918

Note: $\log(6) = \log(2 \times 3) = \log(2) + \log(3) \approx 0.301 + 0.477 = 0.778$

Lecture 2 Quiz Challenge

Test your speed and conversion skills based on Lecture 2 problems.

Question 1 of 5

Score: 0/0

Have Doubts?

Struggling with converting compound units like dynamic pressure or log fractions? Contact Abhishek Sengar Sir to clear up your physics/chemistry foundations!

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