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AP Chemistry Notes

3.4.1 Using PV = nRT to Relate Gas Properties

AP Syllabus focus: ‘For ideal gases, macroscopic properties are related by the ideal gas law, PV = nRT, connecting pressure, volume, temperature, and amount of gas.’

Gases are often modelled with a simple mathematical relationship that links measurable bulk properties. Mastering how to use PV=nRTPV=nRT lets you predict how changing one gas property affects the others under idealized conditions.

The Ideal Gas Model and State Variables

An ideal gas is a simplified model in which gas particles are treated as point-like and in constant random motion, with no attractions and perfectly elastic collisions. In this model, the gas is fully described by four state variables: pressure, volume, temperature, and amount.

Ideal gas law: A relationship for an ideal gas that connects pressure, volume, temperature, and moles through PV=nRTPV=nRT.

This law is used to relate macroscopic measurements (like a lab-measured pressure and volume) to the microscopic count of particles through moles.

Core Equation and Meaning of Each Quantity

The ideal gas law is the central tool for connecting gas properties.

PV=nRT PV = nRT

PP = pressure of the gas (commonly atm, kPa, or Pa)

VV = volume of the gas (commonly L or m3^3)

nn = amount of gas in moles (mol)

RR = ideal gas constant (value depends on units used for PP and VV)

TT = absolute temperature (K)

A key requirement is unit consistency: the numerical value chosen for RR must match the units used for PP and VV, and TT must be in Kelvin.

Temperature must be Kelvin

Because TT is proportional to average particle kinetic energy in this model, temperature is measured on an absolute scale.

  • Convert with T(K)=T(C)+273.15T(\text{K}) = T(^{\circ}\text{C}) + 273.15

  • Never use Celsius directly in PV=nRTPV=nRT

Choosing a compatible value of RR

Common pairings include:

  • R=0.082057 L\cdotpatm\cdotpmol1\cdotpK1R = 0.082057\ \text{L·atm·mol}^{-1}\text{·K}^{-1} for PP in atm and VV in L

  • R=8.314 J\cdotpmol1\cdotpK1R = 8.314\ \text{J·mol}^{-1}\text{·K}^{-1} for energy relationships, noting 1 J=1 Pa\cdotpm31\ \text{J} = 1\ \text{Pa·m}^3

Using PV=nRTPV=nRT to Relate Gas Properties

The strength of PV=nRTPV=nRT is that it directly links how one property must respond when another changes, as long as the sample behaves ideally.

Solving for an unknown state variable

You can rearrange algebraically depending on what is unknown:

  • Moles: n=PVRTn = \dfrac{PV}{RT} (connects measured PP and VV to amount of gas)

  • Pressure: P=nRTVP = \dfrac{nRT}{V} (pressure increases with more moles or higher temperature; decreases with larger volume)

  • Volume: V=nRTPV = \dfrac{nRT}{P} (volume expands as temperature or moles increase; compresses as pressure increases)

  • Temperature: T=PVnRT = \dfrac{PV}{nR} (relates thermal conditions to measured PP and VV for a known amount)

Reasoning about proportional changes (qualitative)

When the amount of gas is fixed, the ideal gas law implies:

  • At constant TT, PP is inversely proportional to VV (compressing a gas raises its pressure)

Pasted image

These graphs illustrate Boyle’s law for a fixed amount of gas at constant temperature. Plotting PP versus VV produces a hyperbola, while plotting 1/P1/P versus VV linearizes the same inverse relationship and makes trends easier to read quantitatively. Together, they reinforce that PVPV remains constant when TT and nn are held constant. Source

  • At constant VV, PP is directly proportional to TT (heating a sealed rigid container raises pressure)

  • At constant PP, VV is directly proportional to TT (heating a balloon increases volume if pressure stays near constant)

When the container conditions are fixed, changing nn has predictable effects:

  • At constant TT and VV, increasing nn increases PP (more particles colliding with the walls)

High-Utility Workflow and Common Pitfalls

Efficient setup steps

  • Identify knowns and unknown; write the target rearranged form first

  • Convert all quantities to compatible units before substituting

  • Track significant figures based on measured inputs

Common errors to avoid

  • Using Celsius instead of Kelvin (systematic and large error)

  • Mixing units (e.g., PP in kPa with RR in L·atm·mol1^{-1}·K1^{-1})

  • Forgetting that VV is the gas volume (not necessarily container volume if gas is collected differently)

FAQ

$R$ is the same constant expressed in different unit systems.

Common forms include:

  • $0.082057\ \text{L·atm·mol}^{-1}\text{·K}^{-1}$

  • $8.314\ \text{Pa·m}^3\text{·mol}^{-1}\text{·K}^{-1}$

Yes.

Rearrange to find moles, then use $M=\dfrac{m}{n}$, where $m$ is the measured mass. This requires accurate $P$, $V$, and $T$ data.

Combine $n=\dfrac{m}{M}$ with $PV=nRT$ to obtain $d=\dfrac{PM}{RT}$, where $d=\dfrac{m}{V}$.

This is useful for comparing densities at different $P$ and $T$.

  • $^\circ\text{C}$ to K: add 273.15

  • L to m$^3$: divide by 1000

  • kPa to Pa: multiply by 1000

It is most reasonable at relatively low pressure and high temperature, where particle volume and attractions have minimal effect on measured $P$ and $V$.

Practice Questions

(2 marks) State the ideal gas law and name the four macroscopic variables it relates.

  • PV=nRTPV=nRT (1)

  • Identifies PP, VV, TT, and nn (1)

(5 marks) A sample of gas has P=95.0 kPaP=95.0\ \text{kPa}, V=2.50 LV=2.50\ \text{L}, and T=35.0CT=35.0^\circ\text{C}. Calculate the amount of gas in moles, using R=8.314 Pa\cdotpm3\cdotpmol1\cdotpK1R=8.314\ \text{Pa·m}^3\text{·mol}^{-1}\text{·K}^{-1}.

  • Converts TT to Kelvin: 308.15 K308.15\ \text{K} (1)

  • Converts PP to Pa: 9.50×104 Pa9.50\times10^4\ \text{Pa} (1)

  • Converts VV to m3^3: 2.50×103 m32.50\times10^{-3}\ \text{m}^3 (1)

  • Uses n=PVRTn=\dfrac{PV}{RT} with correct substitution (1)

  • Correct final nn with appropriate rounding/units (1)

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