Question

How will the velocity of a gas molecule vary if its molecular weight is 1. Under which of the following volumes will 1.00 mol of an ideal gas exhibit the greatest pressure at 300 K?
A. 0.01 L
B. 0.10 L
C. 1.00 L
D. 10.0 L
*Using Graham's law
*Show solution and explain why use that particular formula and why use it.

Answer 1

Velocity of a gas molecule is given by $v_{rms}=\sqrt{\frac{3k_bT}{M}}$

Where M is the molecular mass and $k_b=\frac{R}{N_a}$ is boltzman's constant and $N_a$ is avagadro's number.

Let us take the mass of gas molecule is 1 gram=$10^{-3} kg$

$v_{rms}=\sqrt{\frac{ 3 \times 8.314 \times 300}{1 \times 10^{-3} \times 6.0022 \times 10^{23}}}=$ $1.242 \times 10^{-17} \space m/sec-molecule$

By Grahams law of diffusion,$r(diffusion \space rate)=v \propto \frac{1}{\sqrt{m}}$

If we increase molecular rate,the diffusion speed will decrease.

Under minimum volume will 1.00 mole of an ideal gas exhibit the greatest pressure at 300 K.

Because by Boyle's law,$P \propto \frac{1}{V}$  for constant temperature and a fixed number of moles.

Hence ,For $V=0.01\space ltr$, the pressure will be maximum.

So,option A is correct.