# Ideal Gas Law and dimensional analysis

## Homework Statement

The ideal gas law can be recast in terms of the density of a gas.

a) Use dimensional analysis to find an expression for the density $$\rho$$ of a gas in terms of the number of moles n, the volume V, and the molecular weight M in kilograms per mole.

b) With the expression found in part (a), show that:

P = $$\frac{\rho}{M}$$*RT

for an ideal gas.

c) Find the density of the carbon dioxide atmosphere at the surface of Venus, where the pressure is 90.0 atm and the temperature is 7.00 * 102 K.

d) Would an evacuated steel shell of radius 1.00 m and mass 2.00 * 102 kg rise or fall in such an atmosphere? Why?

## Homework Equations

Ideal Gas Law: PV = nRT

## The Attempt at a Solution

I have never really worked with dimensional analysis, but here's what I have so far:

PV = nRT
[Pa][m3] = [mol][J/mol*K][K]

But I'm not sure how to manipulate this.

Unfortunately, I can't complete the rest of the problem because it's all interdependent, and I'm unsure of how to proceed. Any help would be appreciated.

Related Introductory Physics Homework Help News on Phys.org
You (may) know that

$$n=\frac{m}{M}$$

where m is the mass, M is the molar mass and n is the number of moles. Try substituting this in the ideal gas law and solve for

$$\frac{m}{V}$$

which is the same thing as the density of the gas. Does this help?

You (may) know that

$$n=\frac{m}{M}$$

where m is the mass, M is the molar mass and n is the number of moles. Try substituting this in the ideal gas law and solve for

$$\frac{m}{V}$$

which is the same thing as the density of the gas. Does this help?

Yes.. thank you. I now have proved:

$$\rho$$ = MP/RT

and solved b), showing that the expression is equal to:

($$\rho$$/M)RT = P

However, for part C, I can substitute everything except M. I know

n = m/M and n = PV/RT

but I'm not sure if I'd be applying them correctly to solve for M. Any guidance?

You don't have to substitute M, since you can calculate a value for it. After all it says "carbon dioxide", which has a specific molar mass.

Do you know how to calculate the molar mass for carbon dioxide?

You don't have to substitute M, since you can calculate a value for it. After all it says "carbon dioxide", which has a specific molar mass.

Do you know how to calculate the molar mass for carbon dioxide?
Ah, I see! Now that I have calculated c), I just need some help with d). What exactly is it asking?

What happens to a steel shell (with the specifications you mentioned), if dropped close to the surface of Venus?

There are three options. It will just continue falling down after you've dropped it, until it hits the surface (assuming the surface is solid, I have no idea if it actually is). Perhaps it moves in the opposite direction, upwards. Or maybe it moves in neither of the directions, it stays in the same place.

What happens to a steel shell (with the specifications you mentioned), if dropped close to the surface of Venus?

There are three options. It will just continue falling down after you've dropped it, until it hits the surface (assuming the surface is solid, I have no idea if it actually is). Perhaps it moves in the opposite direction, upwards. Or maybe it moves in neither of the directions, it stays in the same place.

Primarily self-taught, so I'm attempting to work through this problem.

Well, Buoyancy = (density)*(volume)*(gravity)

Would density = density of atmosphere of Venus = 6.81 * 10^(-4) kg/m^3 ?
And volume = volume of evacuated steel shell? How would this be calculated?

Yes you have the right density. However, remember that the acceleration due to gravity has a different value on Venus, so you cannot you g=9.81 m/s^2. (Check the appropriate value from a book or wikipedia).

I have no idea what a steel shell is. I assume that "steel shell" stands for a "steel sphere". After all a radius is given in the text. It's the only thing that makes sence. (I assume that it cannot be a cylinder because we have no value for its height)

Borek
Mentor
Steel sphere, empty inside.

Thank you all for your help!