Solved Examples on Ideal Gas Law
Example 1: What is the volume occupied by 2.34 grams of carbon dioxide gas at STP?
Solution:
Given,
Weight (m) of the carbon dioxide = 2.34 grams
At STP
Temperature = 273.0 K
Pressure = 1.00 atm.R = 0.08206 L atm mol¯1 K¯1
Number of mole n is,
n = m / M
where,
n is the number of moles,
m is the weight
M is the molar mass of the substance.Molar Mass of carbon dioxide = 44.0 g mol¯1
n = 2.34 g / 44.0 g mol¯1
= 0.0532 molAccording to the ideal gas equation,
PV = nRT
V = nRT / P
Substituting all the values,
V = [0.0532) (0.08206) (273.0)] / 1.00
= 1.19 L
Example 2: A sample of argon gas at STP occupies 56.2 litres. Determine the number of moles of argon and the mass of argon in the sample.
Solution:
Given,
Volume (V) of the Argon = 56.2 liters
At STP,
Temperature = 273.0 K.
Pressure = 1.00 atm.Molar mass of Argon gas = 39.948 g/mol
According to Ideal Gas equation,
PV = nRT
n = PV / RT…(1)
Substituting all the values in the above equation,
n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K)]
= 2.50866 mol
Number of mole n,
n = m/M
m = nM…(2)
Substituting all the values in the above equation,
m = (2.50866 mol)×(39.948 g/mol)
= 100 g
Example 3: At what temperature will 0.654 moles of neon gas occupy 12.30 litres at 1.95 atmospheres?
Solution:
Given,
Volume (V) of Neon Gas = 12.30 litres
Pressure = 1.95 atm
Number of Moles = 0.654 moles
According to Ideal Gas Equation,
PV = nRT
T = PV / nR
Substituting all values in above equation,
T = [(1.95 atm) ×(12.30 L)] / [(0.654 mol)×(0.08206 L atm mol¯1 K¯1)]
= 447 K
Example 4: 5.600 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of 300 K. When all the solid CO2 becomes gas, what will be the pressure in the container?
Solution:
Given,
Weight (m) of carbon dioxide = 5.600 g
Volume (V) of the carbon dioxide = 4.00 L
Temperature = 300 K
Molar Mass of carbon dioxide = 44.0 g mol¯1
Number of Mole n,
n = m/M…(1)
Substituting all the values in the above equation,
n = (5.600 g) / (44.009 g/mol)
= 0.1272467 mol
According to the Ideal Gas Equation,
PV = nRT
P = nRT/V…(2)
Substituting all the values in the above equation,
P = (0.1272467 mol)× (0.08206 L atm mol¯1 K¯1)× (300 K)/ (4.00 L)
= 0.7831 atm
Ideal Gas Law
The ideal gas law also called the general gas equation, is an equation that provides the relation among the various parameters of the gas i.e. they provide the relation among pressure(P), temperature(T), and Volume(V) of the gas. It is a combination of Charles’s law, Boyle’s Law, Avogadro’s law, and Gay-Lussac’s law. This law was first stated by the French physicist Benoit Paul Émile Clapeyron in 1834.
Table of Content
- What is Ideal Gas?
- Ideal Gas Laws
- Ideal Gas Law Units
- What is Ideal Gas Equation?
- Equation of Ideal Gas Law
- Derivation of the Ideal Gas Equation
- Ideal Gas Equation Units
- Absolute Temperature
- Relationship between Pressure and Temperature
- Solved Examples on Ideal Gas Law
Let’s learn about Ideal Gas Law and its derivation and others in detail.
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