Ideal Gas Law Q: Increase in Volume w/ 100°C Water -> Steam

[CAF123]

Homework Statement

The ideal gas law states that 1 mole of an ideal gas at 0°C and 1 atmosphere=1.0125x105Pa (standard temperature and pressure, or “STP”) occupies a volume of 22.4litres (2.24x10-2m3).

(i) Use this result to obtain an approximate value for the increase in volume that occurs when a mass of water at 100°C is converted into steam at the same temperature.

For argon (monatomic gas, mass=40amu) at STP, estimate:
(ii) The typical spacing between atoms
(iii) The root-mean-square (rms) speed of the atoms

The Attempt at a Solution

i)I am not sure how to use the ideal gas law STP conditions. For one, this assumes 0 degrees temperature, while the question is concerned with 100 degrees. I know I have to incorporate the latent heat of vapourisation since the water is changing phase. (and at the same temperature)

ii) Assuming small spherical atoms, $r = \left(\frac{6}{\pi n}\right)^{1/3}$, where n = N_a/V, but I think this just takes me in a circle since I don't know V. Also, I think I have to assume ideal gas conditions, so each particle would be of negligible volume.

Many thanks.

 

[Sunil Simha]

Use the Ideal gas equation where the temperature is 373 K. I believe the volume occupied by water in liquid phase is very small compared to that in the gas phase and thus can be neglected in their difference.

To find the spacing between the atoms, you just need to find the average volume occupied by an atom i.e. volume of gas per atom. Though the atom is considered infinitesimally small, you just need to find the volume of its immediate neighborhood and not that of the atom itself.

 

[SteamKing]

In an ideal gas, what is the relationship between pressure, volume, and temperature?

 

[CAF123]

Use the Ideal gas equation where the temperature is 373 K. I believe the volume occupied by water in liquid phase is very small compared to that in the gas phase and thus can be neglected in their difference.

So, when it says use this result, it just means use the ideal gas law? (PV = nRT). (So what I need is V where n=1, T = 373K, P = 1 atm, and simply solve for V?)

To find the spacing between the atoms, you just need to find the average volume occupied by an atom i.e. volume of gas per atom.

So have I not to assume ideal gas conditions? (I.e particles take up negligible volume)

Though the atom is considered infinitesimally small, you just need to find the volume of its immediate neighborhood and not that of the atom itself.

This is what I did, and I derived the eqn in the OP. But n= N_a/V, so this only brings back in V, which is not what I want.

 

[SteamKing]

The OP gives a volume for one mole of gas at STP.

 

[CAF123]

The OP gives a volume for one mole of gas at STP.

Is this in response to my first question?

 

[CAF123]

To find the spacing between the atoms, you just need to find the average volume occupied by an atom i.e. volume of gas per atom. Though the atom is considered infinitesimally small, you just need to find the volume of its immediate neighborhood and not that of the atom itself.

Volume per particle = V/N, V the volume of the container (assume 22.4 l) and $N=N_A$. So the volume per particle is about 3.72 x 10^-23 and so r is then about 2.3 x 10^-16m from the relation given in the OP. Seems a bit small.

 

[Borek]

the volume per particle is about 3.72 x 10^-23

Of what?

and so r is then about 2.3 x 10^-16m

No idea how you got it and what the "relation" from the first post is about. But assuming each atom occupies center of a cube of a volume that you already calculated, it is very easy to calculate the distance. And it is definitely different from the number you posted.

 

[CAF123]

Of what?

The 3.72 x 10^-23l was the volume per particle that I have calculated.

No idea how you got it and what the "relation" from the first post is about.

This is derived by assuming the atoms occupy a volume comprising a sphere.

But assuming each atom occupies center of a cube of a volume that you already calculated, it is very easy to calculate the distance. And it is definitely different from the number you posted.

Assuming this, I get $\approx 3.3 \times 10^{-8}m$.