How Do You Calculate Moles and RMS Speed in Ideal Gas Problems?

  • Thread starter Thread starter Auron87
  • Start date Start date
  • Tags Tags
    Theory
AI Thread Summary
The discussion revolves around calculating moles and root mean square (RMS) speed in ideal gas problems. The user is uncertain about converting gas volume from cm³ to m³, which is crucial for accurate calculations. They correctly convert 160 cm³ to 160 x 10^-6 m³ but mistakenly use the wrong molar mass for neon in their mass calculations. The RMS speed is derived using the formula pV = (1/3)Nm(mean square speed), and the user seeks clarification on the derivation of pressure in relation to gas molecules. The conversation highlights the importance of correct unit conversions and understanding underlying gas laws.
Auron87
Messages
10
Reaction score
0
I have this sheet to do but I think its really hard! I've tried most of the questions but I need someone to tell me if what I've done is right or wrong and what is wrong if it is wrong because I'm not sure on a lot of it!

1. A cylinder contains 2.3 x 10^4 cm3 of an ideal gas at a pressure of 1.8 x 10^7 Pa and a temperature of 27C.

Calculate the amount, in mol, of gas in the cylinder

The main thing here I'm not too sure about is the conversion of the volume into m3. Will it be 2.3 x 10^-2m3? Because this gave me an end answer of 166 moles which seems like quite a lot..

2. Some neon-20 gas, assumed to be ideal, has a volume of 160 cm3 at a pressure of 1.2 x 10^5Pa and a temperature of 23C.

Calculate:
a) the amount of substance in mol

again not really too sure about converting to m3 but this is what I've done:
160 cm3 = 160 x 10^-6m3

pV = nRT
1.2 x 10^5 x 1.6 x 10^-4 = n x 8.31 x 296
n = 7.81 x 10^-3

b) the number of atoms of neon present

Ermm don't know if what is right but I multiplied my answer from a) by Avogadros constant (6.02 x 10^23) to get 4.7 x 10^21.

c) the mass of a neon atom

I used this moles formula here
moles = mass
Mr
mass = 7.81 x 10^-3 x 10 = 7.81 x 10^-2

d) the rms speed of the atoms

Well I guess this depends on whether my previous answers are correct but this is what I did!

pV = 1/3Nm(mean square speed)
1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x (mean square speed)
mean square speed = 73.8
root mean square speed = 8.59 m/s

3. Last question now!

Calculate the factor by which the rms speed of the molecules of an ideal gas will change when temperature of the gas changes from 20C to 120C.

I'm just pretty much stuck with this question!



Thanks very much for any help!
 
Physics news on Phys.org
Question 1 to question 2b seem ok to me.
again not really too sure about converting to m3 but this is what I've done:
160 cm3 = 160 x 10^-6m3
This is correct.

the mass of a neon atom.
I used this moles formula here
moles = mass/Molar mass
mass = 7.81 x 10^-3 x 10 = 7.81 x 10^-2
Molar mass of Neon is 20, not 10

pV = 1/3Nm(mean square speed)
1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x (mean square speed)
pV = 1/3Nm(mean square speed)
1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x mass x (mean square speed)
but not
1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x (mean square speed)

Calculate the factor by which the rms speed of the molecules of an ideal gas will change when temperature of the gas changes from 20C to 120C.
Use the following formula
rms speed = sqrt (3RT/Molar Mass)

I think you should know how to derive the formula above from pV = 1/3Nm(mean square speed)
 
Can someone explain why the pressure due to an ideal gas in a container is given by P = \frac{1}{3}\rho&lt;c^2&gt;? (rho is the density of the gas and <c^2> is the mean square speed)
I happen to have the derivation of the eqn at hand, but somehow, there is this particular step in it I didn't understand; it's the step in which they calculated the change of momentum for a single gas molecule to be 2mv. Since force exerted on container walls is given by F = \frac {\triangle p}{\triangle t} which means the change of momentum (p) divided by the time taken for the momentum to change.

However, in the case of the gas particle, \triangle t was taken to be \frac {2l}{v}! (l is the length of the cuboid container, and v is velocity of the gas molecule). Why?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top