What are the assumptions of the kinetic molecular theory of an ideal gas?

The kinetic molecular theory models an ideal gas as particles in constant random motion, with negligible volume compared with the container, no attractive or repulsive forces between them, and perfectly elastic collisions (no net energy loss). The average kinetic energy of the particles is directly proportional to the Kelvin temperature. Real gases behave most ideally at high temperature and low pressure.

How do you use the combined gas law?

Use $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ from Table T for a fixed mass of gas, with temperatures in Kelvin ($K = {}^\circ\text{C} + 273$). Identify the initial and final conditions, convert any Celsius temperatures, and solve for the unknown. Pressure units cancel as long as they match on both sides. STP from Table A is $273$ K and $101.3$ kPa.

What is the difference between kinetic and potential energy on a heating curve?

On the sloped sections of a heating curve the temperature changes, so the average kinetic energy of the particles changes (within one phase). On the flat plateaus a phase change occurs at constant temperature, so the kinetic energy is unchanged and the added heat increases the potential energy by overcoming the attractions between particles. Temperature change means kinetic energy; a phase-change plateau means potential energy.

When do you use q = mC(delta-T) versus q = mH?

Use $q = mC\Delta T$ when the temperature is changing within a single phase (warming or cooling). Use $q = mH$ for a phase change at constant temperature: $q = mH_f$ for melting or freezing and $q = mH_v$ for boiling or condensing. The water constants (specific heat $4.18\ \text{J/g}\cdot\text{K}$, heat of fusion $334\ \text{J/g}$, heat of vaporization $2260\ \text{J/g}$) are on Table B.

How do you read a solubility curve and calculate molarity?

Table G plots grams of solute per $100$ g of water against temperature; read up from a temperature to the solute's curve and across to the axis to find the saturation amount. Below the curve is unsaturated, on it is saturated, above it is supersaturated. Molarity (Table T) is moles of solute per liter of solution: convert any given mass of solute to moles first, then divide by the volume of solution in liters.

New YorkChemistry

Regents Chemistry physical behavior of matter: a complete skills guide to states, gas laws, heating curves, calorimetry and solutions

A deep-dive Regents Chemistry guide to the physical behavior of matter: states of matter and kinetic molecular theory, the combined gas law with STP, heating and cooling curves and the kinetic-potential energy distinction, calorimetry with the q = mC(delta-T) and q = mH equations, and solutions including solubility curves and concentration, with the Reference Tables and exam technique.

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Jump to a section

Why the physical behavior of matter is the most heavily weighted unit
States of matter and the kinetic molecular theory
The gas laws
Heating curves and energy
Calorimetry
Solutions and concentration
Check your knowledge

Why the physical behavior of matter is the most heavily weighted unit

This unit, states of matter, gases, energy and solutions, contributes a large share of the Regents exam, especially in Part B-2 and Part C, where graph reading and calculations dominate. It also leans on the Reference Tables more than any other unit. This guide ties together the matching dot-point pages, each with its own practice: states of matter and kinetic molecular theory, the gas laws, heating and cooling curves, heat and calorimetry, solutions and solubility curves, and concentration and molarity.

States of matter and the kinetic molecular theory

Solids have fixed, ordered particles that vibrate in place; liquids have close particles that slide past one another; gases have widely spaced particles in rapid random motion. The kinetic molecular theory of an ideal gas assumes constant random motion, negligible particle volume, no intermolecular forces, and elastic collisions. Temperature measures the average kinetic energy of the particles and is directly proportional to the Kelvin temperature.

The gas laws

For a fixed mass of gas, pressure and volume are inversely related at constant temperature, and volume and Kelvin temperature are directly related at constant pressure. The combined gas law unites these:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Always use Kelvin temperature and matching pressure units. STP (Table A) is $273$ K and $101.3$ kPa.

Heating curves and energy

On a heating curve, sloped sections mean the temperature (and so the kinetic energy) is changing within one phase, while flat plateaus mean a phase change at constant temperature, where the added heat raises the potential energy by overcoming particle attractions. Cooling curves are the reverse, with freezing and condensing plateaus where energy is released.

Calorimetry

Choose the equation by whether the temperature changes:

Situation	Equation	Constant (water, Table B)
Warming or cooling (one phase)	$q = mC\Delta T$	$C = 4.18\ \text{J/g}\cdot\text{K}$
Melting or freezing	$q = mH_f$	$H_f = 334\ \text{J/g}$
Boiling or condensing	$q = mH_v$	$H_v = 2260\ \text{J/g}$

A process that releases heat is exothermic; one that absorbs heat is endothermic.

Solutions and concentration

A solution is a homogeneous mixture of solute in solvent. Table G gives saturation amounts (grams per $100$ g water): below the curve is unsaturated, on it saturated, above it supersaturated. Most solids dissolve more at higher temperature; gases dissolve less. Molarity (Table T) is moles of solute per liter of solution; convert a given mass to moles first.

A two-stage heat calculation

How many joules are needed to melt $10.0$ g of ice at $0\,^\circ\text{C}$ and then warm the resulting water to $25.0\,^\circ\text{C}$ ? (Table B: $H_f = 334\ \text{J/g}$ , $C = 4.18\ \text{J/g}\cdot\text{K}$ .)

step 1 Heat to melt the ice (phase change)

$q_1 = mH_f = (10.0)(334) = 3340$ J.

step 2 Heat to warm the water (temperature change)

$\Delta T = 25.0 - 0 = 25.0$ K, so $q_2 = mC\Delta T = (10.0)(4.18)(25.0) = 1045$ J.

step 3 Add the two stages

$q_{\text{total}} = 3340 + 1045 = 4385$ J (about $4.39 \times 10^3$ J).

step 4 Note the method

A phase change uses $q = mH$ and a temperature change uses $q = mC\Delta T$ ; a problem that crosses a plateau needs both, added together.

Solids, liquids and gases differ in particle spacing and motion, and the kinetic molecular theory models an ideal gas as freely moving particles with negligible volume, no forces and elastic collisions, with average kinetic energy proportional to Kelvin temperature. The combined gas law $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ relates a fixed mass of gas using Kelvin temperatures, with STP at $273$ K and $101.3$ kPa. On heating curves, slopes change kinetic energy and plateaus change potential energy during phase changes. Use $q = mC\Delta T$ for temperature changes and $q = mH$ for phase changes, with water constants from Table B. Solutions are classified by Table G as unsaturated, saturated or supersaturated, and molarity is moles of solute per liter of solution.

Check your knowledge

Attempt these under timed conditions, then check the solutions.

State two assumptions of the kinetic molecular theory of an ideal gas. (2 marks)
A gas at $2.0$ L and $300$ K is heated to $600$ K at constant pressure. Find the new volume. (2 marks)
On a heating curve, state what is happening to the energy during the boiling plateau. (1 mark)
Calculate the heat needed to warm $25.0$ g of water by $10.0$ K ( $C = 4.18\ \text{J/g}\cdot\text{K}$ ). (2 marks)
Using a Table G value of about $40$ g per $100$ g water at a temperature, classify a solution holding $25$ g per $100$ g water at that temperature. (1 mark)
Calculate the molarity of $0.50$ mol of solute in $0.25$ L of solution. (2 marks)

Solutions

Q1: Any two of: particles are in constant random motion; their volume is negligible; there are no forces between them; collisions are elastic.
Q2: Constant pressure, so $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ : $V_2 = 2.0 \times \dfrac{600}{300} = 4.0$ L.
Q3: The temperature (and kinetic energy) is constant; the added heat increases the potential energy by overcoming the attractions between particles.
Q4: $q = mC\Delta T = (25.0)(4.18)(10.0) = 1045$ J (about $1.05 \times 10^3$ J).
Q5: $25$ g is below the saturation value of about $40$ g, so the solution is unsaturated.
Q6: $\text{molarity} = \dfrac{0.50}{0.25} = 2.0$ M.

Sources & how we know this

Physical Setting/Chemistry Core Curriculum — New York State Education Department (2002)
Reference Tables for Physical Setting/Chemistry, 2011 Edition — New York State Education Department (2011)