recall and use the equation of state for an ideal gas expressed as pV = nRT, where n = amount of substance (number of moles) and as pV = NkT, where N = number of molecules

Equation of State for an Ideal Gas

Learning Objective

Recall and use the equation of state for an ideal gas expressed as

\$pV = nRT\$

where n is the amount of substance (moles), and also as

\$pV = NkT\$

where N is the total number of molecules.

Key Concepts

• Pressure (p) – force per unit area exerted by the gas on the container walls.
• Volume (V) – the space occupied by the gas.
• Temperature (T) – a measure of the average kinetic energy of the molecules (in kelvin).
• Amount of substance (n) – number of moles, related to the number of molecules by N = nNA, where NA = 6.022×10^{23} mol^{-1} (Avogadro’s constant).
• Universal gas constant (R) – \$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\$.
• Boltzmann constant (k) – \$k = 1.381\times10^{-23}\ \text{J K}^{-1}\$.

Deriving the Two Forms

1. Start from the mole‑based form: \$pV = nRT\$.
2. Replace the amount of substance with the number of molecules: \$n = \dfrac{N}{N_A}\$.
3. Substitute into the equation:

   \$pV = \left(\frac{N}{NA}\right)RT = N\left(\frac{R}{NA}\right)T.\$

4. Recognise that \$\displaystyle \frac{R}{N_A}=k\$, the Boltzmann constant, giving the molecule‑based form:

   \$pV = NkT.\$

Comparison of \cdot ariables and Constants

Using the Equation of State

Typical rearrangements allow you to solve for any one variable when the others are known:

• Pressure: \$p = \dfrac{nRT}{V} = \dfrac{NkT}{V}\$
• Volume: \$V = \dfrac{nRT}{p} = \dfrac{NkT}{p}\$
• Temperature: \$T = \dfrac{pV}{nR} = \dfrac{pV}{Nk}\$
• Moles: \$n = \dfrac{pV}{RT}\$
• Molecules: \$N = \dfrac{pV}{kT}\$

Worked Example

Problem: A container holds \$2.5\times10^{23}\$ molecules of an ideal gas at a pressure of \$1.2\times10^{5}\ \text{Pa}\$ and a temperature of \$300\ \text{K}\$. Find the volume of the gas.

Solution:

1. Use the molecule‑based form: \$pV = NkT\$.
2. Rearrange for \$V\$: \$V = \dfrac{NkT}{p}\$.
3. Insert the numbers:

   \$V = \frac{(2.5\times10^{23})(1.381\times10^{-23}\ \text{J K}^{-1})(300\ \text{K})}{1.2\times10^{5}\ \text{Pa}}.\$

4. Calculate:

   \$\$V \approx \frac{(2.5)(1.381)(300)}{1.2}\times10^{-5}\ \text{m}^{3}

   \approx 8.6\times10^{-3}\ \text{m}^{3} \; (8.6\ \text{L}).\$\$

Common Pitfalls

• Confusing \$n\$ (moles) with \$N\$ (molecules). Remember \$N = nN_A\$.
• Using temperature in °C instead of kelvin. Convert with \$T(\text{K}) = T(^{\circ}\text{C}) + 273.15\$.
• Mixing units: ensure pressure is in pascals, volume in cubic metres, and constants in compatible units.
• Assuming the ideal‑gas law holds at very high pressures or very low temperatures where real‑gas behaviour becomes significant.

Summary

The ideal‑gas equation of state links pressure, volume, temperature, and the amount of gas. It can be written in terms of moles (\$pV=nRT\$) or in terms of individual molecules (\$pV=NkT\$). Mastery of the equation allows rapid calculation of any one variable when the others are known, provided the gas behaves ideally.

Practice Questions

1. Calculate the pressure exerted by \$0.500\ \text{mol}\$ of an ideal gas occupying \$12.0\ \text{L}\$ at \$298\ \text{K}\$.
2. A sample contains \$3.0\times10^{22}\$ molecules of an ideal gas at \$1.00\ \text{atm}\$ and \$350\ \text{K}\$. Find its volume in \$\text{cm}^{3}\$.
3. How many moles of gas are required to fill a \$5.00\ \text{L}\$ container at \$2.00\ \text{atm}\$ and \$273\ \text{K}\$?
4. Explain why the ideal‑gas equation may fail for a gas at \$0.1\ \text{K}\$ and \$100\ \text{atm}\$.