Q.We have 0.5 g of hydrogen gas in a cubic chamber of size 3cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius 1 Å).

Concept: Compare the molecules' own volume with the volume the ideal gas law predicts for the compressed state.

1. Moles: $n = 0.5\ \text{g}/2\ \text{g/mol} = 0.25\ \text{mol}$.
2. Ideal-law final volume (Boyle's law, $V_1=27\ \text{cm}^3$ at $P_1=1\ \text{atm}$, $P_2=100\ \text{atm}$): $V_2 = 27/100 = 0.27\ \text{cm}^3$.
3. Actual molecular volume: each H$_2$ molecule ($r=1\times10^{-8}\ \text{cm}$) has $V_{\text{molecule}}\approx4.19\times10^{-24}\ \text{cm}^3$; total for $\approx1.5\times10^{23}$ molecules is $\approx0.63\ \text{cm}^3$. …

Why this comparison is the right test

The ideal gas law $pV=nRT$ assumes the volume of the gas molecules themselves is negligible next to the container's volume. That assumption is safe at low pressure but breaks down when the container is squeezed down toward the molecules' own size. The test: compute the volume the gas would occupy under the ideal gas law in the compressed state, and compare it with the actual volume the molecules physically take up.

Step-by-step

1. Moles of hydrogen gas.

Molar mass of H$_2$ is $2\ \text{g/mol}$, and we have $0.5\ \text{g}$:

2. Volume the gas would occupy under the ideal gas law, in the final state.

$V_1 = 3^3 = 27\ \text{cm}^3$ at $P_1=1\ \text{atm}$ (NTP). Compressed isothermally to $P_2=100\ \text{atm}$, Boyle's law gives:

3. Actual volume occupied by the hydrogen molecules themselves.

Each H$_2$ molecule is a sphere of radius $r = 1\ \text{\AA} = 1\times10^{-8}\ \text{cm}$:

With $n = 0.25\ \text{mol}$ ($N_A = 6.022\times10^{23}\ \text{mol}^{-1}$), total number of molecules $\approx 1.5\times10^{23}$, so:

4. Compare. …

Alternate approach — packing-fraction shortcut. Instead of separately computing both volumes and comparing them, estimate the packing fraction directly: $\phi = \dfrac{N V_{molecule}}{V_2}$. Using $N\approx1.5\times10^{23}$ molecules, $V_{molecule}\approx4.2\times10^{-24}\text{ cm}^3$, and $V_2 = 0.27\text{ cm}^3$, $\phi \approx 2.3$ — the molecules alone would need over twice the available volume. As a rule of thumb, the ideal-gas a …