Question

Helium gas is in a cylinder that has rigid walls. If the pressure of the gas is 2.00 atm, then the root-mean-square speed of the helium atoms is vrms = 176 m/s. By how much (in atmospheres) must the pressure be increased to increase the vrms of the He atoms by 100 m/s? Ignore any change in the volume of the cylinder

Answer 1

Answer:

$p=p_{2}-p_{1}=2.9188atm$

Explanation:

First step

To calculate by how much the pressure has been increased we must find a relation between Vrms and the pressure p

$V_{rms}=\sqrt{\frac{3RT}{M} }$

Where M is molar mass,R is gas constant and T is temperature

As R and M is constant so we obtain

Vrams∝√T

Second step

From Ideal gas law we know that

$pV=nRT$

we will find that

p∝T

So from first and second step, we can obtain that

Vrms∝√p

And a relation between both of them could be given by:

$\frac{V_{rms1} }{V_{rms2}}=\frac{\sqrt{p_{1} } }{\sqrt{p_{2}} }\\ p_{2}=[\frac{(V_{rms1})^{2} }{(V_{rms2})^{2} }]p_{1}\\ p_{2}=\frac{(276m/s)^{2} }{(176m/s)^{2} } (2atm)\\p_{2}=4.9188atm$

The pressure is increased to 4.1988 atm, so the amount will be given by:

$p=p_{2}-p_{1}=4.9188atm-2atm\\p=p_{2}-p_{1}=2.9188atm$