Ideal Gas Equation and Real Gases
Derivation of the Ideal Gas Equation
The Ideal Gas Equation is obtained by combining Boyle's Law, Charles's Law, and Avogadro's Law into a single relationship. Boyle's law states that volume is inversely proportional to pressure when temperature and moles are constant ($V \propto \frac{1}{P}$). Charles's law states that volume is directly proportional to temperature when pressure and moles are constant ($V \propto T$). Avogadro's law states that volume is directly proportional to the number of moles when pressure and temperature are constant ($V \propto n$). When none of the variables are held constant, all three relationships merge:
$$V \propto \frac{nT}{P}$$
Combining the proportions: Replacing the proportionality sign with a constant gives $V = \text{Constant} \times \frac{nT}{P}$. The constant is the Universal Gas Constant, denoted by $R$.
Rearranging: Multiplying both sides by $P$ gives $PV = nRT$ — the ideal gas equation. It states that for any quantity of an ideal gas, the product of pressure and volume equals the product of moles, the gas constant, and absolute temperature.
Reduction to individual laws: When $T$ and $n$ are constant, $PV = nRT$ becomes $PV = k$ (Boyle's Law). When $P$ and $n$ are constant, it becomes $V = kT$ (Charles's Law). When $P$ and $T$ are constant, it becomes $V = kn$ (Avogadro's Law).
For practical calculations where a fixed sample of gas changes from one set of conditions ($P_1$, $V_1$, $T_1$) to another ($P_2$, $V_2$, $T_2$), the ideal gas equation takes a convenient ratio form. For one mole of gas, $\frac{PV}{T} = R$, which is constant. Therefore:
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
When to use: This combined form is used whenever a gas undergoes a change in two or more of its parameters (pressure, volume, temperature) and the amount of gas remains constant.
Unit consistency: Always convert temperature to Kelvin and ensure pressure and volume use compatible units on both sides of the equation.
Density form: By substituting $n = \frac{m}{M}$ (where $m$ is mass and $M$ is molar mass), the equation becomes $PV = \frac{m}{M}RT$. Rearranging gives $d = \frac{PM}{RT}$, where $d$ is the density of the gas. This shows density is directly proportional to molar mass and pressure, and inversely proportional to temperature.
The Universal Gas Constant R
The value of $R$ is calculated using the fact that one mole of an ideal gas occupies 22.414 dm³ at STP (1 atm pressure, 273.16 K). Substituting these into $R = \frac{PV}{nT}$:
$$R = \frac{1 \text{ atm} \times 22.414 \text{ dm}^3}{1 \text{ mol} \times 273.16 \text{ K}} = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$$
R is a universal constant whose value depends only on the units chosen for pressure, volume, and temperature. Its physical meaning: one mole of ideal gas absorbs 0.0821 dm³·atm of energy when heated by 1 K at constant pressure.
$R$=Universal gas constant — same for all ideal gases(varies with unit system)
$P$ in atm, $V$ in dm³
→$R = 0.0821$ dm³ atm K⁻¹ mol⁻¹
$P$ in torr, $V$ in dm³
→$R = 62.4$ dm³ torr K⁻¹ mol⁻¹
$P$ in torr, $V$ in cm³
→$R = 62400$ cm³ torr K⁻¹ mol⁻¹
$P$ in Nm⁻², $V$ in m³
→$R = 8.314$ J K⁻¹ mol⁻¹
Energy in calories
→$R = 1.989$ cal K⁻¹ mol⁻¹
Atmosphere to torr conversion: $1 \text{ atm} = 760 \text{ torr}$, so $R = 0.0821 \times 760 = 62.4$ dm³ torr K⁻¹ mol⁻¹. Since $1 \text{ mm Hg} = 1 \text{ torr}$, these units are interchangeable.
dm³ to cm³ conversion: $1 \text{ dm}^3 = 1000 \text{ cm}^3$, so $R = 62.4 \times 1000 = 62400$ cm³ torr K⁻¹ mol⁻¹.
SI calculation: Using $P = 101325 \text{ Nm}^{-2}$, $V = 0.022414 \text{ m}^3$, $T = 273.16 \text{ K}$, and $1 \text{ Nm} = 1 \text{ J}$, we get $R = 8.314$ J K⁻¹ mol⁻¹.
Calorie conversion: Since $1 \text{ cal} = 4.18 \text{ J}$, $R = \frac{8.314}{4.18} = 1.989$ cal K⁻¹ mol⁻¹.
Common Values of R
•
$0.0821$ dm³ atm K⁻¹ mol⁻¹ (most common in chemistry problems)
•
$62.4$ dm³ torr K⁻¹ mol⁻¹ (when pressure is in torr/mm Hg)
•
$62400$ cm³ torr K⁻¹ mol⁻¹ (when volume is in cm³)
•
$8.314$ J K⁻¹ mol⁻¹ (SI units)
•
$1.989$ cal K⁻¹ mol⁻¹ (energy in calories)
Ideal vs Real Gases
An Ideal Gas is a theoretical gas that perfectly obeys all gas laws under all conditions of temperature and pressure. Its behaviour is described by two key postulates of the Kinetic Molecular Theory: there are no forces of attraction between gas molecules, and the actual volume of gas molecules is negligible compared to the volume of the container. A Real Gas, on the other hand, shows measurable deviations from ideal behaviour because these assumptions break down under certain conditions.
No intermolecular forces (ideal): Ideal gas molecules do not attract or repel each other. Real gas molecules do experience intermolecular forces — the strength depends on the nature of the gas.
Negligible molecular volume (ideal): Ideal gas molecules are treated as point masses with zero volume. Real gas molecules occupy a finite, non-negligible volume, especially at high pressure.
Compressibility factor: The ratio $\frac{PV}{nRT}$ equals exactly 1 for an ideal gas at all conditions. For real gases, this ratio deviates from 1 — it is called the Compressibility Factor. A value above 1 means the gas is less compressible than ideal; below 1 means more compressible.
Ideal vs Real Gas Behaviour
•
Ideal: $PV = nRT$ holds at all $P$ and $T$ — Real: deviates at high $P$ and low $T$
•
Ideal: no intermolecular forces — Real: attractive forces reduce observed pressure
•
Ideal: zero molecular volume — Real: molecules have finite volume (excluded volume $b$)
•
Ideal: compressibility factor $Z = 1$ always — Real: $Z$ varies with $P$ and $T$
•
Ideal: obeys Boyle's, Charles's, and Avogadro's laws exactly — Real: deviates especially at extreme conditions
Experimental observations with gases like $H_2$, $He$, $N_2$, and $CO_2$ reveal two clear trends about when real gases behave ideally and when they do not. The Compressibility Factor graphs ($PV/RT$ vs pressure) show deviations that shift with temperature.
Low pressure = ideal behaviour: At low pressures, gas molecules are far apart, intermolecular forces are negligible, and molecular volume is insignificant compared to the container — the gas behaves ideally.
High pressure = non-ideal behaviour: At high pressures, molecules are forced close together. Intermolecular attractive forces become significant and the finite volume of molecules can no longer be ignored — the gas deviates from ideality.
High temperature = ideal behaviour: At elevated temperatures, molecules move fast enough that intermolecular attractions are overcome by kinetic energy. At 100°C, deviations are much smaller than at 0°C for the same gas.
Low temperature = non-ideal behaviour: At low temperatures, molecules move slower and intermolecular forces have a greater relative effect, causing larger deviations from the ideal gas equation.
van der Waals Equation for Real Gases
van der Waals (1873) identified the two postulates of the kinetic molecular theory that fail for real gases. He introduced corrections for both pressure and volume to modify the ideal gas equation into an equation that describes real gas behaviour. The two causes of deviation from ideality are intermolecular attractive forces and the finite volume of gas molecules.
Pressure correction: A molecule striking the container wall is pulled inward by neighbouring molecules, reducing its impact. The observed pressure $P$ is therefore less than the ideal (kinetic) pressure $P_i$. To compensate, a correction term $\frac{a}{V^2}$ is added, where $a$ measures the strength of intermolecular attraction.
Volume correction: Real gas molecules occupy space, so the volume available for free movement is less than the container volume. An exclusion term $b$ (the Excluded Volume) is subtracted from the measured volume $V$. The excluded volume is approximately four times the actual volume of the molecules: $b = 4V_m$.
Nature of constants: The constant $a$ reflects the strength of intermolecular forces — gases with stronger attractions (like $SO_2$, $Cl_2$) have higher $a$ values. The constant $b$ reflects the size of molecules — larger molecules have higher $b$ values.
van der Waals Constants for Common Gases
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$H_2$: $a = 0.245$ atm dm⁶ mol⁻², $b = 0.0266$ dm³ mol⁻¹
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$O_2$: $a = 1.360$ atm dm⁶ mol⁻², $b = 0.0318$ dm³ mol⁻¹
•
$N_2$: $a = 1.390$ atm dm⁶ mol⁻², $b = 0.0391$ dm³ mol⁻¹
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$CO_2$: $a = 3.590$ atm dm⁶ mol⁻², $b = 0.0428$ dm³ mol⁻¹
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$NH_3$: $a = 4.170$ atm dm⁶ mol⁻², $b = 0.0371$ dm³ mol⁻¹
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$SO_2$: $a = 6.170$ atm dm⁶ mol⁻², $b = 0.0564$ dm³ mol⁻¹
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$Cl_2$: $a = 6.493$ atm dm⁶ mol⁻², $b = 0.0562$ dm³ mol⁻¹
Applying both corrections to the ideal gas equation $PV = nRT$ gives the van der Waals Equation. For one mole of gas, the corrected pressure $(P + \frac{a}{V^2})$ replaces $P$ and the corrected volume $(V - b)$ replaces $V$:
$$(P + \frac{a}{V^2})(V - b) = RT$$
This equation modifies the ideal gas law to account for intermolecular forces (pressure correction) and molecular volume (volume correction) in real gases.
$P$=Observed (measured) pressure of the real gas(atm or Nm⁻²)
$a$=van der Waals constant for intermolecular attraction — larger for gases with stronger forces(atm dm⁶ mol⁻² or Nm⁴ mol⁻²)
$V$=Measured volume of the container(dm³ or m³)
$b$=Excluded volume — the volume occupied by 1 mole of molecules in compressed state (≈ 4 × actual molecular volume)(dm³ mol⁻¹ or m³ mol⁻¹)
$R$=Universal gas constant(matches units of P and V)
$T$=Absolute temperature(K)
$n$ moles of gas (not 1 mole)
→$(P + \frac{n^2a}{V^2})(V - nb) = nRT$
$a = 0$ and $b = 0$
→Reduces to the ideal gas equation $PV = nRT$
Derivation of pressure correction: The inward pull on a wall-striking molecule is proportional to the concentration of gas molecules ($\propto n/V$). Since both the striking molecule and the pulling molecules contribute, the correction is proportional to $(n/V)^2$. Thus $P' = \frac{an^2}{V^2}$, and for one mole, $P' = \frac{a}{V^2}$.
For n moles: The full van der Waals equation for $n$ moles is $(P + \frac{n^2a}{V^2})(V - nb) = nRT$. Note that the pressure correction scales with $n^2$ while the volume correction scales with $n$.
Application Strategy
Worked Examples
Problem Statement
A sample of nitrogen gas is enclosed in a vessel of volume 380 cm³ at 120 °C and pressure of 101325 Nm⁻². The gas is transferred to a 10 dm³ flask and cooled to 27 °C. Calculate the pressure exerted by the gas at 27 °C.
ideal-gas-ratio-form
Strategic Toolkit
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
Step 1
Convert all quantities to consistent units
Convert volume from cm³ to dm³ and temperature from Celsius to Kelvin.
$$V_1 = 380 \text{ cm}^3 = 0.38 \text{ dm}^3, \quad T_1 = 120 + 273 = 393 \text{ K}, \quad T_2 = 27 + 273 = 300 \text{ K}$$
Step 2
Identify known values
List the four known quantities and the unknown.
$$P_1 = 101325 \text{ Nm}^{-2}, \quad V_1 = 0.38 \text{ dm}^3, \quad T_1 = 393 \text{ K}, \quad V_2 = 10 \text{ dm}^3, \quad T_2 = 300 \text{ K}$$
Step 3
Rearrange the combined gas equation
Solve for the final pressure P₂ by isolating it on one side.
$$P_2 = \frac{P_1V_1T_2}{T_1V_2}$$
Step 4
Substitute and calculate
Plug in all values and compute the result.
$$P_2 = \frac{101325 \times 0.38 \times 300}{393 \times 10} = 2938.4 \text{ Nm}^{-2}$$
Problem Statement
Calculate the density of CH₄ gas at 0 °C and 1 atm. What will the density be if (a) the temperature is increased to 27 °C, and (b) the pressure is increased to 2 atm at 0 °C?
gas-density
Strategic Toolkit
$$d = \frac{PM}{RT}$$
Step 1
Calculate density at 0 °C and 1 atm
Convert temperature to Kelvin and substitute all values into the density formula. Molar mass of CH₄ is 16 g mol⁻¹.
$$d = \frac{1 \times 16}{0.0821 \times 273} = 0.7138 \text{ g dm}^{-3}$$
Step 2
Density at 27 °C (part a)
Temperature increases to 300 K, which causes the gas to expand and density to decrease.
$$d = \frac{1 \times 16}{0.0821 \times 300} = 0.649 \text{ g dm}^{-3}$$
Step 3
Density at 2 atm and 0 °C (part b)
Doubling the pressure compresses the gas, approximately doubling the density.
$$d = \frac{2 \times 16}{0.0821 \times 273} = 1.427 \text{ g dm}^{-3}$$
Problem Statement
Calculate the mass of 1 dm³ of NH₃ gas at 30 °C and 1000 mm Hg pressure, assuming ideal behaviour.
mass-from-ideal-gas
Strategic Toolkit
$$m = \frac{PVM}{RT}$$
Step 1
Convert pressure to atm and temperature to Kelvin
Pressure must be converted from mm Hg to atm by dividing by 760. Temperature must be in Kelvin.
$$P = \frac{1000}{760} = 1.315 \text{ atm}, \quad T = 30 + 273 = 303 \text{ K}$$
Step 2
Apply the mass form of the ideal gas equation
Rearrange $PV = \frac{m}{M}RT$ to solve for mass. Molar mass of NH₃ is 17 g mol⁻¹.
$$m = \frac{PVM}{RT} = \frac{1.315 \times 1 \times 17}{0.0821 \times 303}$$
Step 3
Compute the result
Evaluate the numerical expression to find the mass of 1 dm³ of NH₃ gas under these conditions.
$$m = 0.907 \text{ g}$$
Problem Statement
One mole of methane gas is maintained at 300 K in a volume of 250 cm³. Calculate the pressure exerted by the gas (i) assuming ideal behaviour, and (ii) assuming non-ideal behaviour using the van der Waals equation. Given: $a = 2.253$ atm dm⁶ mol⁻², $b = 0.0428$ dm³ mol⁻¹.
ideal-pressurenon-ideal-pressure
Strategic Toolkit
$$P = \frac{nRT}{V} \quad \text{(ideal)} \qquad P = \frac{nRT}{V - nb} - \frac{n^2a}{V^2} \quad \text{(van der Waals)}$$
Step 1
Convert volume and identify known values
Convert 250 cm³ to dm³ and list all parameters.
$$V = 250 \text{ cm}^3 = 0.25 \text{ dm}^3, \quad n = 1, \quad T = 300 \text{ K}, \quad R = 0.0821$$
Step 2
Calculate ideal pressure
Apply $PV = nRT$ directly.
$$P_{ideal} = \frac{1 \times 0.0821 \times 300}{0.25} = 98.5 \text{ atm}$$
Step 3
Calculate non-ideal pressure using van der Waals
Apply the van der Waals equation rearranged for pressure. The volume correction reduces the denominator and the attraction correction subtracts from the result.
$$P = \frac{1 \times 0.0821 \times 300}{0.25 - 0.0428} - \frac{1 \times 2.253}{(0.25)^2} = \frac{24.63}{0.2072} - \frac{2.253}{0.0625} = 118.91 - 36.05 = 82.86 \text{ atm}$$
Step 4
Compare the two pressures
The non-ideal pressure is 82.86 atm, which is about 15.64 atm less than the ideal prediction. This difference shows the gas is non-ideal at these conditions — intermolecular attractions reduce the observed pressure significantly.
$$\Delta P = 98.5 - 82.86 = 15.64 \text{ atm}$$
Problem Statement
Calculate the value of the universal gas constant R in SI units (J K⁻¹ mol⁻¹) using the fact that one mole of an ideal gas occupies 22.414 dm³ at STP (1 atm, 273.16 K). Given: 1 atm = 101325 Nm⁻², 1 Nm = 1 J.
gas-constant-si
Strategic Toolkit
$$R = \frac{PV}{nT}$$
Step 1
Convert all quantities to SI units
Pressure must be in Nm⁻² and volume in m³.
$$P = 101325 \text{ Nm}^{-2}, \quad V = 0.022414 \text{ m}^3, \quad n = 1 \text{ mol}, \quad T = 273.16 \text{ K}$$
Step 2
Substitute into the formula for R
Apply $R = \frac{PV}{nT}$ and simplify using 1 Nm = 1 J.
$$R = \frac{101325 \times 0.022414}{1 \times 273.16} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$$
Memorize formulas
Ideal Gas Equation
Relates P, V, n, R, T. Use when any gas property needs to be calculated from the others. R = 0.0821 dm³ atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹.
Formula
$$PV = nRT$$
Combined Gas Equation (Ratio Form)
Use when a fixed amount of gas changes between two states. No R or n needed — all units must be consistent on both sides.
Formula
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
Gas Density Formula
Relates density to molar mass, pressure, and temperature. d increases with M and P, decreases with T. Use R = 0.0821 when P is in atm and V in dm³.
Formula
$$d = \frac{PM}{RT}$$
Gas Constant (Atmosphere Units)
Use when pressure is in atm and volume is in dm³. Most common in chemistry textbook problems.
Formula
$$R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$$
Gas Constant (SI Units)
Use when pressure is in Nm⁻² (Pa) and volume is in m³.
Formula
$$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$$
van der Waals Equation (1 mole)
Corrects ideal gas law for real gases. 'a' = attraction constant (atm dm⁶ mol⁻²), 'b' = excluded volume (dm³ mol⁻¹). Use when gas is at high pressure or low temperature.
Formula
$$(P + \frac{a}{V^2})(V - b) = RT$$
van der Waals Equation (n moles)
For n moles of real gas. Pressure correction scales with n², volume correction scales with n.
Formula
$$(P + \frac{n^2a}{V^2})(V - nb) = nRT$$
Review
1
The ideal gas equation $PV = nRT$ can be derived by combining which three gas laws?
Ans
Boyle's law, Charles's law, and Avogadro's law
Review
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Boyle's Law
At constant temperature and moles, pressure and volume are inversely proportional
$$PV = k_1 \text{ (constant } T, n\text{)}$$
Charles's Law
At constant pressure and moles, volume is directly proportional to temperature
$$\frac{V}{T} = k_2 \text{ (constant } P, n\text{)}$$
Avogadro's Law
At constant temperature and pressure, volume is directly proportional to moles
$$\frac{V}{n} = k_3 \text{ (constant } T, P\text{)}$$
Combine All Three
Combining the proportionalities gives the ideal gas equation
$$PV = nRT$$
2
A sample of oxygen gas occupies 10.0 dm³ at 2.0 atm pressure and 300 K. How many moles of O₂ are present? (Use $R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$)
Ans
0.812 mol
Review
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Rearrange the Equation
Solve the ideal gas equation for the number of moles
$$n = \frac{PV}{RT}$$
Substitute Values
Insert $P = 2.0$ atm, $V = 10.0$ dm³, $R = 0.0821$, $T = 300$ K
$$n = \frac{2.0 \times 10.0}{0.0821 \times 300}$$
Calculate
Evaluate the numerator and denominator
$$n = \frac{20.0}{24.63} = 0.812 \text{ mol}$$
3
A gas at 1.0 atm and 273 K occupies 22.4 dm³. If the pressure is changed to 2.5 atm and the temperature to 546 K, what is the new volume?
Ans
17.9 dm³
Review
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Combined Gas Law
Write the ratio form of the combined gas equation
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
Solve for V₂
Rearrange to isolate the unknown volume
$$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$$
Substitute and Calculate
Insert the given values and evaluate
$$V_2 = \frac{1.0 \times 22.4 \times 546}{273 \times 2.5} = 17.9 \text{ dm}^3$$
4
What is the value of the universal gas constant $R$ in units of dm³ atm K⁻¹ mol⁻¹?
Ans
0.0821
Review
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Gas Constant in atm Units
The numerical value of $R$ depends on the chosen units for pressure and volume
$$R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$$
SI Equivalent
The same constant in SI units has a different numerical value
$$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$$
5
Which value of $R$ should be used when pressure is measured in mmHg and volume in dm³?
Ans
62.36 dm³ mmHg K⁻¹ mol⁻¹
Review
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Match Units
$R$ must be consistent with the units of $P$, $V$, and $T$ in $PV = nRT$
$$R = 62.36 \text{ dm}^3 \text{ mmHg K}^{-1} \text{ mol}^{-1}$$
Common R Values
Each pressure unit has a corresponding $R$ value
$$0.0821 \text{ (atm)}, \quad 62.36 \text{ (mmHg)}, \quad 8.314 \text{ (Pa m}^3\text{)}$$
6
At STP, 1.0 mol of an ideal gas occupies 22.4 dm³ at 273 K and 1.0 atm. Using the ideal gas equation, calculate the value of $R$:
Ans
0.0821 dm³ atm K⁻¹ mol⁻¹
Review
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Rearrange for R
Solve the ideal gas equation for the gas constant
$$R = \frac{PV}{nT}$$
Substitute STP Values
Insert $P = 1.0$ atm, $V = 22.4$ dm³, $n = 1.0$ mol, $T = 273$ K
$$R = \frac{1.0 \times 22.4}{1.0 \times 273}$$
Calculate
Evaluate to find $R$
$$R = \frac{22.4}{273} = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$$
7
Which statement is NOT an assumption of the kinetic molecular theory of gases?
Ans
All gas molecules move with the same speed at a given temperature
Review
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Review KMT Postulates
The kinetic molecular theory makes several simplifying assumptions about ideal gas behavior
$$\text{Five postulates govern ideal gas behavior}$$
Speed Distribution
Molecules follow a Maxwell-Boltzmann speed distribution — a range of speeds, not a single value
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \quad \text{but individual speeds vary}$$
8
Under which set of conditions does a real gas deviate most significantly from ideal behavior?
Ans
High pressure and low temperature
Review
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Effect of High Pressure
At high pressure, molecules are forced close together, so excluded volume becomes significant
$$\text{Molecular volume can no longer be neglected}$$
Effect of Low Temperature
At low temperature, molecules have less kinetic energy and intermolecular forces become dominant
$$\text{Intermolecular attractions reduce effective pressure}$$
Combined Effect
Both conditions simultaneously maximize non-ideal behavior
$$\text{Maximum deviation at high } P \text{ and low } T$$
9
For a real gas, the compressibility factor $Z = \frac{PV}{nRT}$ exceeds 1 at very high pressures. The primary reason is:
Ans
The finite volume of molecules reduces available space, increasing effective pressure
Review
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Compressibility Factor
For an ideal gas, $Z = 1$ always. Real gases have $Z > 1$ or $Z < 1$ depending on conditions
$$Z = \frac{PV}{nRT}$$
Moderate Pressures
At moderate pressures, intermolecular attractions reduce effective pressure
$$Z < 1 \text{ (attractions dominate)}$$
Very High Pressures
At very high pressures, excluded volume effects dominate, increasing effective pressure
$$Z > 1 \text{ (excluded volume dominates)}$$
10
In the van der Waals equation $(P + \frac{a}{V^2})(V - b) = RT$, the constant $a$ corrects for:
Ans
Intermolecular attractive forces between molecules
Review
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Pressure Correction
Real gas pressure is reduced by intermolecular attractions, so a positive correction is added
$$P_{\text{ideal}} = P_{\text{measured}} + \frac{a}{V^2}$$
Volume Correction
The constant $b$ (separate from $a$) corrects for the finite molecular volume
$$V_{\text{ideal}} = V_{\text{measured}} - b$$
11
The density of an ideal gas is given by $d = \frac{PM}{RT}$. According to this formula, at constant temperature and pressure, the density of a gas is:
Ans
Directly proportional to its molar mass
Review
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Density Formula
Starting from the ideal gas equation, derive the density relationship
$$d = \frac{PM}{RT}$$
Constant Conditions
At constant $T$ and $P$, all terms except $M$ are constants
$$d \propto M \quad (\text{at constant } T \text{ and } P)$$
12
What mass of ammonia gas (NH₃, M = 17 g mol⁻¹) will occupy 5.0 dm³ at 3.0 atm and 300 K? (Use $R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$)
Ans
10.4 g
Review
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Find Moles
Rearrange the ideal gas equation to solve for moles
$$n = \frac{PV}{RT} = \frac{3.0 \times 5.0}{0.0821 \times 300} = \frac{15.0}{24.63} = 0.609 \text{ mol}$$
Convert to Mass
Multiply moles by the molar mass of NH₃
$$m = n \times M = 0.609 \times 17 = 10.4 \text{ g}$$
13
At room temperature and moderate pressure, which of the following gases would deviate most from ideal behavior?
Ans
Ammonia
Review
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Factors Affecting Deviation
Deviation increases with stronger intermolecular forces and larger molecular size
$$\text{Stronger IMF} \rightarrow \text{larger } a \rightarrow \text{greater deviation}$$
Compare the Gases
NH₃ has hydrogen bonding (strongest among options); noble gases have only London forces
$$\text{NH}_3 \text{ (hydrogen bonding) deviates most}$$
14
For 1 mol of CO₂ (a = 3.59 dm⁶ atm mol⁻², b = 0.0427 dm³ mol⁻¹) confined in 5.0 dm³ at 300 K, the pressure calculated using the van der Waals equation is closest to:
Ans
4.82 atm
Review
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Rearrange van der Waals Equation
Solve for pressure from $(P + a/V^2)(V - b) = RT$
$$P = \frac{RT}{V - b} - \frac{a}{V^2}$$
Calculate Corrected Pressure Term
Compute $RT/(V - b)$ with $V - b = 5.0 - 0.0427 = 4.957$ dm³
$$\frac{0.0821 \times 300}{4.957} = \frac{24.63}{4.957} = 4.97 \text{ atm}$$
Subtract Attraction Correction
Subtract $a/V^2 = 3.59/25.0 = 0.144$ atm
$$P = 4.97 - 0.144 = 4.82 \text{ atm}$$
15
At STP (0°C and 1 atm), one mole of any ideal gas occupies approximately:
Ans
22.4 dm³
Review
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Apply Ideal Gas Equation
Set $n = 1$ mol, $P = 1$ atm, $T = 273$ K and solve for volume
$$V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273}{1}$$
Calculate Molar Volume
Evaluate to find the standard molar volume
$$V = 22.4 \text{ dm}^3$$
16
In the van der Waals equation, the constant $b$ represents the excluded volume per mole. For gas molecules modeled as hard spheres, the excluded volume is related to the actual volume of the molecules by:
Ans
$b = 4V_{\text{molecules}}$
Review
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Molecular Volume
Volume of one spherical molecule of radius $r$
$$V_{\text{mol}} = \frac{4}{3}\pi r^3$$
Excluded Volume Per Pair
Two molecules cannot approach closer than $2r$
$$V_{\text{excluded, pair}} = \frac{4}{3}\pi(2r)^3 = 8V_{\text{mol}}$$
Per Molecule (Shared)
Dividing by 2 since the excluded volume is shared between two molecules
$$\frac{V_{\text{excluded, pair}}}{2} = 4V_{\text{mol}}$$
Result
The excluded volume constant $b$ equals four times the total molecular volume per mole
$$b = 4V_{\text{molecules}}$$
17
A real gas behaves most like an ideal gas under which conditions?
Ans
Low pressure and high temperature
Review
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Low Pressure Effect
At low pressure, the space between molecules is large compared to molecular size
$$\text{Volume effects negligible}$$
High Temperature Effect
At high temperature, molecular kinetic energy overcomes intermolecular attractions
$$\text{Intermolecular forces negligible}$$
Combined
Both conditions together approximate ideal gas behavior
$$\text{Real gas } \approx \text{ ideal gas at low } P \text{ and high } T$$
18
The van der Waals equation for $n$ moles of a gas is:
Ans
$(P + \frac{n^2a}{V^2})(V - nb) = nRT$
Review
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Pressure Correction
The number of intermolecular interacting pairs scales with $n^2$
$$P_{\text{corr}} = \frac{n^2 a}{V^2}$$
Volume Correction
The excluded volume is additive across all $n$ moles
$$V_{\text{corr}} = nb$$
Full Equation
Combining corrections with $nRT$ on the right side
$$(P + \frac{n^2a}{V^2})(V - nb) = nRT$$
19
For a real gas at moderate pressure, the pressure predicted by the ideal gas equation is typically higher than the actual pressure. This is because:
Ans
Intermolecular attractive forces reduce the pressure exerted on the container walls
Review
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Ideal Gas Assumption
The ideal gas equation assumes no intermolecular forces between molecules
$$P_{\text{ideal}} = \frac{nRT}{V}$$
Real Gas Effect
Intermolecular attractions reduce the momentum of molecules hitting the walls
$$P_{\text{real}} < P_{\text{ideal}} \text{ at moderate pressures}$$
van der Waals Correction
The constant $a$ quantifies the attractive force correction
$$P_{\text{corrected}} = P + \frac{n^2a}{V^2}$$
20
The density of methane (CH₄, M = 16 g mol⁻¹) at 2.0 atm and 300 K is: (Use $R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$)
Ans
1.30 g dm⁻³
Review
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Density Formula
The density of a gas relates pressure, molar mass, and temperature
$$d = \frac{PM}{RT}$$
Substitute Values
Insert $P = 2.0$ atm, $M = 16$ g mol⁻¹, $R = 0.0821$, $T = 300$ K
$$d = \frac{2.0 \times 16}{0.0821 \times 300}$$
Calculate
Evaluate to find the density
$$d = \frac{32}{24.63} = 1.30 \text{ g dm}^{-3}$$
Question 1 of 20
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1
The ideal gas equation $PV = nRT$ can be derived by combining which three gas laws?