We have $0.5 \, g$ of hydrogen gas in a cubic chamber of size $3 \, cm$ kept at $NTP$. The gas in the chamber is compressed keeping the temperature constant until a final pressure of $100 \, atm$ is reached. Is one justified in assuming the ideal gas law in the final state? (Hydrogen molecules can be considered as spheres of radius $1 \, \mathring{A}$).

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(B) The ideal gas law is valid only if the volume occupied by the gas molecules is negligible compared to the total volume of the container.
$1$. Calculate the number of moles of $H_2$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.5 \, g}{2 \, g/mol} = 0.25 \, mol$.
$2$. Calculate the total number of $H_2$ molecules $(N)$: $N = n \times N_A = 0.25 \times 6.023 \times 10^{23} \approx 1.506 \times 10^{23}$ molecules.
$3$. Calculate the volume of a single $H_2$ molecule $(v_m)$: $v_m = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (10^{-10} \, m)^3 \approx 4.19 \times 10^{-30} \, m^3$.
$4$. Calculate the total volume occupied by the molecules $(V_{mol})$: $V_{mol} = N \times v_m = 1.506 \times 10^{23} \times 4.19 \times 10^{-30} \approx 6.31 \times 10^{-7} \, m^3$.
$5$. Calculate the final volume of the chamber $(V_f)$ using Boyle's Law $(P_i V_i = P_f V_f)$:
$V_i = (3 \, cm)^3 = 27 \, cm^3 = 27 \times 10^{-6} \, m^3$.
$P_i = 1 \, atm$,$P_f = 100 \, atm$.
$V_f = \frac{P_i V_i}{P_f} = \frac{1 \times 27 \times 10^{-6}}{100} = 2.7 \times 10^{-7} \, m^3$.
$6$. Comparison: The volume occupied by the molecules $(6.31 \times 10^{-7} \, m^3)$ is greater than the total volume of the container $(2.7 \times 10^{-7} \, m^3)$. Since the molecular volume is not negligible,the ideal gas law is $NOT$ justified.

Explore More

Similar Questions

Gases begin to conduct electricity at low pressure because

What is the effect of a decrease in the number of molecules per unit volume on the mean free path?

The mean free path for a gas,with molecular diameter $d$ and number density $n$,can be expressed as:

Estimate the mean free path for a water molecule in water vapour at $373 \; K$. The density of water is $1000 \; kg \; m^{-3}$. The density of water vapour at $100 \; ^{\circ}C$ and $1 \; atm$ pressure is $0.6 \; kg \; m^{-3}$. The volume of a molecule multiplied by the total number gives,what is called,molecular volume.

The mean free path for a gas at temperature $300 \ K$ and pressure $600 \ \text{torr}$ is $10^{-7} \ m$. The mean free path of the gas at a temperature $400 \ K$ and pressure $200 \ \text{torr}$ will be

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo