Understanding Molar Volume and Graphical Relationships in Gases
- You’re inflating a balloon for a party.
- With each breath, it grows larger, and you might start to wonder: how much space does the air inside actually take up?
- What if you could measure the volume of the gas and count the number of molecules inside?
This curiosity leads us to an important concept in chemistry: molar volume.
What is Molar Volume?
Molar volume
The molar volume of a gas is the volume occupied by one mole of an ideal gas under specific conditions of temperature and pressure.
- This property is a cornerstone of gas behavior, as described by the ideal gas law.
- At STP (Standard Temperature and Pressure):
- Temperature = 273.15 K (0°C)
- Pressure = 100 kPa
- Molar Volume = 22.7 dm³ mol⁻¹
- In simple terms, this means that one mole of any ideal gas will occupy 22.7 dm³ of space under these conditions, regardless of its chemical identity.
Suppose you have 2 moles of oxygen gas (O₂) at STP. What is the total volume of the gas?
Solution
- Using the molar volume at STP:$$ \text{Volume} = \text{Moles} \times \text{Molar Volume}$$ $$\text{Volume} = 2 \, \text{mol} \times 22.7 \, \text{dm}^3 \ \text{mol}^{-1} = 45.4 \, \text{dm}^3$$
- Thus, 2 moles of oxygen gas occupy 45.4 dm³ at STP.
- Always confirm that the temperature and pressure match STP conditions when using the molar volume of 22.7 dm³ mol⁻¹.
- If the conditions differ, use the ideal gas equation to calculate the volume.
Graphical Relationships Between Temperature, Pressure, and Volume
- The behavior of gases can be visualized through graphs that illustrate the relationships between key variables: pressure (p), volume (V), and temperature (T).
- These relationships are governed by the gas laws, which are derived from the ideal gas equation: $$pV = nRT$$
- Let’s explore these relationships one by one.
Pressure and Volume (Boyle’s Law)
- Consider squeezing a balloon.
- As you reduce its volume, you feel the pressure inside increase: this illustrates Boyle’s Law.
Boyle's law
Boyle's law states that at constant temperature and for a fixed amount of gas, pressure is inversely proportional to volume.
- Mathematically: $$
p \propto \frac{1}{V} \quad \text{or} \quad pV = \text{constant} $$ - Graphically:
- A graph of p vs. V forms a downward-sloping curve.
- A graph of p vs. 1/V forms a straight line.
A gas occupies 4.0 dm³ at a pressure of 100 kPa. If the volume is reduced to 2.0 dm³, what will the new pressure be (assuming constant temperature)?
Solution
Using Boyle’s Law:
$$
p_1V_1 = p_2V_2
$$
$$
100 \, \text{kPa} \times 4.0 \, \text{dm}^3 = p_2 \times 2.0 \, \text{dm}^3
$$
$$
p_2 = \frac{100 \times 4.0}{2.0} = 200 \, \text{kPa}
$$
The pressure doubles to 200 kPa when the volume is halved.
- Many students forget to keep the temperature constant when applying Boyle’s Law.
- Always ensure that no temperature change occurs during the process.
Volume and Temperature (Charles’s Law)
- Now imagine heating a balloon.
- As the temperature rises, the balloon expands: this demonstrates Charles’s Law.
Charles's law
Charles's law states that at constant pressure and for a fixed amount of gas, volume is directly proportional to absolute temperature (in Kelvin).
- Mathematically: $$V \propto T \quad \text{or} \quad \frac{V}{T} = \text{constant}$$
- Graphically:
- A graph of V vs. T (in Kelvin)produces a straight line passing through the origin.
A gas occupies 3.0 dm³ at 273 K. What will its volume be at 546 K (assuming constant pressure)?
Solution
Using Charles’s Law:
$$
\frac{V_1}{T_1} = \frac{V_2}{T_2}
$$
$$
\frac{3.0 \, \text{dm}^3}{273 \, \text{K}} = \frac{V_2}{546 \, \text{K}}
$$
$$
V_2 = \frac{3.0 \times 546}{273} = 6.0 \, \text{dm}^3}
$$
The volume doubles as the temperature doubles (in Kelvin).
- Always convert temperature to Kelvin when using Charles’s Law.
- The Kelvin scale starts at absolute zero, where the volume of a gas theoretically becomes zero.
Pressure and Temperature (Gay-Lussac’s Law)
- Finally, consider a sealed can of soda left in the sun.
- As the temperature rises, the pressure inside increases: this is explained by Gay-Lussac’s Law.
Gay-Lussac's law
Gay-Lussac's law states that at constant volume and for a fixed amount of gas, pressure is directly proportional to absolute temperature (in Kelvin).
- Mathematically:
$$
p \propto T \quad \text{or} \quad \frac{p}{T} = \text{constant}
$$ - Graphically:
- A graph of p vs. T (in Kelvin) produces a straight line passing through the origin.
Gay-Lussac’s Law in Action
A gas has a pressure of 150 kPa at 300 K. What will the pressure be at 450 K (assuming constant volume)?
Solution
Using Gay-Lussac’s Law:
$$
\frac{p_1}{T_1} = \frac{p_2}{T_2}
$$
$$
\frac{150 \, \text{kPa}}{300 \, \text{K}} = \frac{p_2}{450 \, \text{K}}
$$
$$
p_2 = \frac{150 \times 450}{300} = 225 \, \text{kPa}
$$
The pressure increases as the temperature increases.
Gas Laws
- Why is it important to use Kelvin instead of Celsius when studying gas laws?
- How would the graphical relationships change for a real gas under high pressure or low temperature?
- Can you think of any situations where deviations from ideal gas behavior might have significant consequences?
