Answer to Question #93775 in General Chemistry for Sumicad

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.

Please show step by step from the previous question Question #93669
, how to get V=0.01 ltr, all of them step by step on how to get the answer not just the description , and the use of Root-Mean-Square velocities of gaseous particles, step by step instructions please

Answer 1

Velocity of a gaseous molecule$=$ $v_{rms}=\sqrt{\frac{3k_bT}{M}}$

$M=Molecular\ mass$

Molecular mass is equal to $1\frac{g}{mol}=0.001\frac{kg}{mol}$

putting the values in formula

$v_{rms}=\sqrt{\frac{3\times 1.38\times 10^{-23}\times T}{0.001}}=2.0347\times 10^{-10}\sqrt{T}\frac{m}{sec} \ \ \ \ \ \ \ \ \ .........(1)$

Equation $(1)\$ gives the variation of of velocity of gaseous molecule with molecular weight 1 with temprature.

Morever , According to graham's law

rate of diffusion $∝$ $\sqrt{\frac{1}{M}}$

or $v$ $∝\sqrt{\frac{1}{M}}$

using ideal gas equation

$PV=nRT$

$P=\frac{nRT}{V}$

​  $T=300\ K,R=8.314,\ n=1$

$PV=2494.2\ J\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ P=\frac{2494.2}{V}$

$P ∝\frac{1}{V}$

So pressure will be maximum when volume is minimum

So pressure will be maximum when volume is $0.01\ l$