Ideal Gas Law Notes

Ch. 14: Ideal Gas Law

The Volume Amount

Equal volumes of gases at the same temperature (T) and pressure (P) contain the same number of molecules. Volume (V) is proportional to the amount (number of molecules/moles) for gases at the same T and P.

The Ideal Gas Law

The Ideal Gas Law is represented by the equation:
PV=nRTPV = nRT
where:

  • P is the pressure

  • V is the volume

  • n is the number of moles

  • R is the Ideal Gas constant

  • T is the temperature

The value of R depends on the units of P:

  • R=0.08206atmLKmoleR = 0.08206 \frac{atm \cdot L}{K \cdot mole}

  • R=62.4torrLKmoleR = 62.4 \frac{torr \cdot L}{K \cdot mole}

  • R=0.083145barLKmoleR = 0.083145 \frac{bar \cdot L}{K \cdot mole}

STP Conditions

STP stands for Standard Temperature and Pressure.

  • Standard Temperature: 273.15 K (0 °C)

  • Standard Pressure: 1 bar (0.987 atm) - IUPAC standard. Note that previously, 1 atm was often used as standard pressure.

Combined Gas Laws

The combined gas laws can be derived from the ideal gas law. Given:
P<em>1V</em>1=n<em>1RT</em>1P<em>1V</em>1 = n<em>1RT</em>1
P<em>2V</em>2=n<em>2RT</em>2P<em>2V</em>2 = n<em>2RT</em>2

Depending on the variables held constant, various gas laws can be derived. Some examples:

  • P<em>1P</em>2=V<em>2V</em>1\frac{P<em>1}{P</em>2} = \frac{V<em>2}{V</em>1} (Boyle's Law - constant T and n)

  • V<em>1V</em>2=T<em>1T</em>2\frac{V<em>1}{V</em>2} = \frac{T<em>1}{T</em>2} (Charles's Law - constant P and n)

  • P<em>1P</em>2=T<em>1T</em>2\frac{P<em>1}{P</em>2} = \frac{T<em>1}{T</em>2} (Gay-Lussac's Law - constant V and n)

  • P<em>1V</em>1T<em>1=P</em>2V<em>2T</em>2\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2} (Combined Gas Law - constant n)

  • P<em>1V</em>1n<em>1RT</em>1=1\frac{P<em>1V</em>1}{n<em>1RT</em>1} = 1

  • P<em>2V</em>2n<em>2RT</em>2=1\frac{P<em>2V</em>2}{n<em>2RT</em>2} = 1

  • P<em>1V</em>1n<em>1T</em>1=P<em>2V</em>2n<em>2T</em>2\frac{P<em>1V</em>1}{n<em>1T</em>1} = \frac{P<em>2V</em>2}{n<em>2T</em>2} (General Gas Law)

Ideal Gas Law Problems

To solve Ideal Gas Law problems, isolate the needed variable and plug in the values. For example:

  • Calculate the pressure (in bar) of 3.5 moles of He which occupies 20.00 L at 100.0 °C.

  • How many moles of fluorine occupy 9.8 L at 95.33 °C and 800.0 torr?

Molar Mass (MM)

When molar mass (MM) or grams appear in a problem, use a modified form of the Ideal Gas Equation:

MM=mnMM = \frac{m}{n}

Where mm is mass in grams and nn is number of moles.

By replacing nn in the Ideal Gas Equation, PV=nRTPV = nRT

PV=mRTMMPV = \frac{mRT}{MM}

MM=mRTPVMM = \frac{mRT}{PV}

Calculating Molar Mass of a Gas

Example problems:

  • What is the molar mass of an unknown gas if 5.25 g occupies 8.25 L at 1.50 bar and 600.00 K?

  • If 5.00 g of helium take up 3.50 L at 1.33 atm, find the temperature of the gas in °C.

Density of a Gas

Density is defined as mass/volume:

D=mVD = \frac{m}{V}

Using the equation for Molar Mass:

MM=mRTPVMM = \frac{mRT}{PV}

MM=DRTPMM = \frac{DRT}{P}

D=MMPRTD = \frac{MM \cdot P}{RT}

Example problem:

  • What is the density of krypton at 50.00 °C and 1500.0 torr?

Calculating Density of a Gas

Example problem:

  • The density of an unknown gas is 1.66 g/L at 293.5 K and 2.00 bar. What noble gas is it?

Molar Volume

Molar volume is the volume occupied by one mole of a gas.

V=nRTPV = \frac{nRT}{P}

Vn=RTP\frac{V}{n} = \frac{RT}{P}

Find the molar volume of a gas at STP.

Stoichiometry: Ideal Gas Equation Method

Example problem:

Ammonia gas is formed from its elements:

a) Write a balanced equation.

b) How many liters of ammonia form from 3.00 g nitrogen at STP?

c) How many liters of ammonia form from 5.50 g hydrogen at 2.50 atm and 450.0 K?

Stoichiometry: Ideal Gas Equation Method (continued)

d) How many grams of hydrogen are used when 6.80 L nitrogen are used at 650.0 torr and 50.00 °C?

The balanced equation is:

3H<em>2(g)+N</em>2(g)2NH3(g)3 H<em>2(g) + N</em>2(g) \rightarrow 2 NH_3(g)

Equations to Keep Handy

  • When two sets of conditions are mentioned: P<em>1V</em>1T<em>1=P</em>2V<em>2T</em>2\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}

  • Ideal Gas Equation: PV=nRTPV = nRT

  • When molar mass or grams are mentioned: MM=mRTPVMM = \frac{mRT}{PV}

  • When density is mentioned: MM=DRTPMM = \frac{DRT}{P}

  • For molar volume: Vn=RTP\frac{V}{n} = \frac{RT}{P}

  • For stoichiometry problems, if V, P, and T are given for a gas, calculate moles using the ideal gas equation first.

  • If V is wanted in the problem, first do stoichiometry, then use the ideal gas equation to calculate V for the desired gas.

Stoichiometry Pathways

grams A \rightarrow moles A \rightarrow moles B \rightarrow grams B
(V, M) (V, M)

(P,V,T)A \rightarrow (P,V,T)B
NA NA
units A units B
atoms atoms
ions ions
molecules molecules
formula units formula units

Examples

Given the reaction:

SO<em>2(g)+CaCO</em>3(s)+O<em>2(g)CaSO</em>4(s)+CO2(g)SO<em>2(g) + CaCO</em>3(s) + O<em>2(g) \rightarrow CaSO</em>4(s) + CO_2(g)

a. 42.7 L of sulfur dioxide at 543 torr and 22.0 °C react with 33.8 L of oxygen at 287 K and 1.44 atm. How many dg of calcium sulfate can form?
b. How many grams of sulfur dioxide and oxygen are left at the end of the reaction?