Ideal gases

Temperature Scales

Learning Objective

Understand how temperature is measured, convert between common scales, and apply the absolute temperature scale (Kelvin) in the ideal‑gas equation.

Key Temperature Scales

• Celsius (°C) – based on the freezing (0 °C) and boiling point (100 °C) of water at 1 atm.
• Kelvin (K) – the absolute temperature scale; 0 K is absolute zero. The size of one kelvin is identical to one degree Celsius.
• Fahrenheit (°F) – used mainly in the United States; water freezes at 32 °F and boils at 212 °F.

Conversion Between Scales

Quantity	°C	K	°F
Freezing point of water	0	273.15	32
Boiling point of water	100	373.15	212
Absolute zero	-273.15	0	-459.67

General conversion formulas:

• From Celsius to Kelvin: $T(\mathrm{K}) = T(^{\circ}\mathrm{C}) + 273.15$
• From Kelvin to Celsius: $T(^{\circ}\mathrm{C}) = T(\mathrm{K}) - 273.15$
• From Celsius to Fahrenheit: $T(^{\circ}\mathrm{F}) = \frac{9}{5}\,T(^{\circ}\mathrm{C}) + 32$
• From Fahrenheit to Celsius: $T(^{\circ}\mathrm{C}) = \frac{5}{9}\,\bigl[T(^{\circ}\mathrm{F}) - 32\bigr]$

Why Kelvin is Required for Ideal Gases

The ideal‑gas law is expressed as

$$pV = nRT$$

where $p$ is pressure, $V$ is volume, $n$ is the amount of substance (in moles), $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ is the universal gas constant, and $T$ must be in kelvin. Using any other temperature scale would give incorrect numerical results because the law is derived from kinetic theory, which assumes an absolute temperature proportional to the average kinetic energy of the molecules:

$$\langle E_{\text{kin}}\rangle = \frac{3}{2}k_{\mathrm{B}}T$$

Here $k_{\mathrm{B}}$ is Boltzmann’s constant. The proportionality holds only when $T$ is measured from absolute zero.

Practical Example

1. Calculate the pressure exerted by 2.00 mol of an ideal gas confined in a 5.00 L container at 25 °C.
2. Convert the temperature to kelvin: $$T = 25 + 273.15 = 298.15\ \mathrm{K}$$
3. Use the ideal‑gas equation: $$p = \frac{nRT}{V} = \frac{(2.00\ \text{mol})(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})(298.15\ \mathrm{K})}{5.00\ \mathrm{L}}$$

   Convert the volume to cubic metres (1 L = $1.0\times10^{-3}\ \mathrm{m^{3}}$): $$V = 5.00\times10^{-3}\ \mathrm{m^{3}}$$

   $$p = \frac{(2.00)(8.314)(298.15)}{5.00\times10^{-3}} = 9.9\times10^{5}\ \mathrm{Pa}$$
4. Result: $p \approx 9.9\times10^{5}\ \mathrm{Pa}$ (≈ 9.8 atm).

Common Mistakes to Avoid

• Using Celsius or Fahrenheit directly in the ideal‑gas equation.
• Neglecting to convert the volume to SI units (cubic metres) when using $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
• Assuming the gas constant $R$ has the same numerical value in all unit systems; always check the units.