What is the final translational rms speed of the atoms

[stellies]

Hi, I have a thermodynamics problem which I tried to solve but have no idea whether my attempt has been succesful. Here is the problem:

Initially the translational root mean squared speed of an atom of a monatomic
ideal gas is 250 m/s. The pressure and volume of the gas are each doubled while
the number of moles of the gas is kept constant. What is the final translational rms
speed of the atoms.

I used the equation: v(rms) = sqrt[(3RT)/M].

Therefore:

250 = sqrt[(3RT)/M]

I then squared both sides and took over the M (molar mass):

(3RT) = 62500M

and then divided both sides by 3:

RT = 20833(1/3)M

I then argued that becuase PV=nRT

RT = PV/n

= PVM/m (from n = m/M)

I then had the expression:

PVM/m = 20833(1/3)M

and eliminated M on both sides.

PV/m = 20833(1/3)

Therefore, by doubling the Pressure, P and Volume V:

2PV/m = 41666(2/3).

From here I just worked backwards.

2(PV/m) = 41666(2/3)

= 2(RT/M) = 41666(2/3)

and eventually got back to:

sqrt[(3RT)/M] = 250

I obviously worked in a circle...

So if anybody could give me a hint where I can get out of this circle...

That would be greatly appreciated.

Thanks

 

[Kurdt]

When you came to multiply PV by 2, you did it to both sides, so you've effectively done nothing. You don't have to manipulate the equation all that much, you could have replaced the RT in the original equation and eliminated all the manipulation.

 

[alphysicist]

Hi stellies,

Thanks for the PM. Kurdt's suggestion is definitely the best way to go; once you write it the way he says you can solve it in just a few lines of work.

Your approach would work, but here are the places your solution went wrong:

Hi, I have a thermodynamics problem which I tried to solve but have no idea whether my attempt has been succesful. Here is the problem:

Initially the translational root mean squared speed of an atom of a monatomic
ideal gas is 250 m/s. The pressure and volume of the gas are each doubled while
the number of moles of the gas is kept constant. What is the final translational rms
speed of the atoms.

I used the equation: v(rms) = sqrt[(3RT)/M].

Therefore:

250 = sqrt[(3RT)/M]

I then squared both sides and took over the M (molar mass):

(3RT) = 62500M

and then divided both sides by 3:

RT = 20833(1/3)M

This line is not correct; you have divided the left side by 3, but you have divided the right side by 9.

I then argued that becuase PV=nRT

RT = PV/n

= PVM/m (from n = m/M)

I then had the expression:

PVM/m = 20833(1/3)M

and eliminated M on both sides.

PV/m = 20833(1/3)

Therefore, by doubling the Pressure, P and Volume V:

2PV/m = 41666(2/3).

Of course Kurdty already pointed out that this is not right. The action you have done to the right side is correct (since it is being multiplied by an overall factor of 4); however, on the left side you want P and V to be the new pressure and volume, so the factor of 2 should not be there.


That should give you the answer, but definitely try it the other way. You have two equations: PV = n RT and v_{\rm rms} = \sqrt{3 RT/M} (and also the definition of n). Put those together to get one equation without T, and you'll have just one equation to work with.

 

[stellies]

Hi thanks for the replies.

But I still can't seem to get my head around this one.

Could you please elaborate on the last statement:

You have two equations: PV = nRT and V(rms) = sqrt(3RT/M). (and also the definition of n). Put those together to get one equation without T, and you'll have just one equation to work with.

Thanks

 

[Kurdt]

Well what does RT equal from the ideal gas equation? Then plug that into the rms equation.

 

[stellies]

Hi thanks for sticking out for me...

So after sticking in RT = PV/n into v(rms) = sqrt(3RT/M) I get:

v(rms) = sqrt(3PV/m).

250 = sqrt(3PV/m).

Now somehow I need to get an expression where P and V is doubled..and I can't seem to get to that point :-(

...

if

62500 = 3PV/m

which one should I double, left or right?...?

sorry but I seem to be getting nowhere, no matter what I try...another hint would be appreciated

Thanks

 

[stellies]

Hi me again...I tried to work it out and I got an answer of 353.55 m/s...(if allowed) is there any chance that that may be right ?

 

[Kurdt]

You've missed the M out in the previous manipulations. So now you know that if you double pressure and volume you will get some other rms velocity. If you divide the equations the variables will all drop out and you'll be left with your unknown and some numbers.

 

[Kurdt]

Hi me again...I tried to work it out and I got an answer of 353.55 m/s...(if allowed) is there any chance that that may be right ?

Pressure and volume are each doubled which means we have 2V x 2P = 4PV.

 

[stellies]

Right, thanks I really didn't see that last part (4PV)...

Anyway...

Here is another shot at it:

250 = sqrt(3RT/M), therefore

250 = sqrt(3PV/m)

62500m = 3PV

By doubling P AND V (4PV)

62500m 12PV

and then

15625 = 3PV/m <=> 15625 = 3RT/M <=> 125 = sqrt(3RT/M)

therefore my new v(rms) speed is 125 m/s

Looks strange to me..I probably went wrong again... :-(

If I did, what do you mean with

"If you divide the equations the variables will all drop out and you'll be left with your unknown and some numbers"

 

[Kurdt]

The problem comes with the doubled volume and pressure equation. You don't know the rms velocity so you can't say its 62500. Its just an unknown variable. By that time you should have 2 equations which you can divide. The first equation where you know the rms velocity and have the pressure and volume variables, and another equation with the doubled variables and an unknown rms velocity. If you divide the two equations all the variables drop out and you're left with the unknown rms velocity and a bunch of numbers you can rearrange to get the answer.

 

[stellies]

Hi

I think I finally got it. The new rms must be 500 m/s . .. . . . ?

Has to be.. If it is, it is actually simple and I actually feel silly that I was so blind.

But thanks a lot for your help and sticking it out

 

[Kurdt]

That looks fine to me.