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Matters and Properties of Matters
Atomic Structure with Examples
Periodic Table
The Mole Concept with Examples
Gases with Examples
--Measuring Pressure of Gas and Manometers with Examples
--Kinetic Theory of Gases
--Effusion and Diffusion of Gases with Examples
--Gas Laws with Examples
--Ideal Gas Law with Examples
--Density of Gases with Examples
--Mixtures and Partial Pressure of Gases with Examples
--Gases cheat Sheet
Chemical Reactions with Examples
Nuclear Chemistry (Radioactivity)
Acids and Bases
Rates of Reactions (Chemical Kinetics)
Chemical Equilibrium
Chemical Bonds
Exams and Problem Solutions
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Ideal Gas Law with Examples

Ideal Gas Law

n ideal gas; molecules have no volume and there are no interaction between them. In real there is no such a gas, it is just an assumption. All real gases has small volumes and there are interactions between them. In problem solutions; we assume all gases as ideal gas. Given equation below is ideal gas law. We get it by combining all gas laws given in last section.


Where; P pressure, V volume, n number of particles, R gas constant 0,08206 L atm / K mol or 22,4/273 L atm / K mol, and T temperature

Now we solve some problems related to ideal gas law for better understanding, follow each example carefully.

Example: Find pressure of 8,8 g CO2 at 27 0C in container having volume 1230 cm3. (C=12, O=16)

Solution: We first find molar mass of CO2;


Then, we find moles of CO2;

n=8,8/44=0,2 moles

Converting temperature from 0C to K and volume from cm3 to liter;

T=27+273=300 K

V=1230 cm3=1,23 liters

Now, we use ideal gas law to find unknown quantity.



P=4 atm

Example: Find molar mass of X(gas) given in the picture below having volume 896 cm3, temperature 273 0C and mass 0,96g. (O=16, and atmospheric pressure is 1 atm )



We make unit conversions first.

P=38 cm Hg=38/76=0,5 atm

V=896/1000=0,896 liters

T=273 + 273=546 K

Now, we use ideal gas law to find n;



n=0,03 moles

Molar mass of X;


MX=32 g/mol

Thus; X(gas)=O2

Example: System given below is placed in a location having 70 cm Hg atmospheric pressure. Container has 2 g He at first, then we add 1 g H2 gas to this container. Find the rising of Hg in the manometer after adding H2. (He=4, H2=2)

Pinitial=P0-20=70-20=50 cm Hg

ninitial=2/4=0,5mol He

nH2=1/2=0,5mol H2



We write ideal gas law for initial and final values, then we dive them each other to find unknown value.

V and T are constant in two situations.

Pfinal=100 cm Hg

Difference between levels of Hg between two branches of manometer;

100-70=30 cm in right branch of manometer

Thus difference between initial and final levels of Hg becomes;

20+30=50 cm

This change is shared by two branches of manometer ;

50/2=25 cm

Thus, Hg rises 25 cm in one branch of manometer.

Gases Exams and  Problem Solutions

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