# Ideal-gas Equation

## Homework Statement

A tank contains 25.0 kg of O2 gas at a gauge pressure of 9.10 atm. If the oxygen is replaced by fluorine, how many
kilograms of the latter will be needed to produce a gauge pressure of 4.00 atm?

PV=nRT

## The Attempt at a Solution

For the O2 we know P, can find n (m=nM), and R is a constant. Still leaves one equation with two unknowns? Am I to assume some temperature, find the volume of the container, and then use that same temperature to calculate the mass of flourine. Is that valid?

it's been a while since thermo but can't you negate the temperature and volume of the container because they are fixed? then you are left with the knowns and unknowns you need to solve? then it's P_1/n_1=P_2/n_2 right? forgive me if i'm way off.

it's been a while since thermo but can't you negate the temperature and volume of the container because they are fixed? then you are left with the knowns and unknowns you need to solve? then it's P_1/n_1=P_2/n_2 right? forgive me if i'm way off.

I have no clue.

try breaking the problem into two parts:

if you wanted to maintain original pressure, how many kg of Flourine needed?

Hint no of moles must be the same, tho you don't need to calculate that figure.

and for rest of problem, since PV=nRT where V and T are same, R is constant you have relation where P is directly proportional to n.

see above post..

Still leaves one equation with two unknowns? Am I to assume some temperature, find the volume of the container, and then use that same temperature to calculate the mass of flourine. Is that valid?
Yes. The volume is the same obviously, and the pressure, and you can assume the same temperature without loss of generality. So you can find n because R is always the same.

try breaking the problem into two parts:

if you wanted to maintain original pressure, how many kg of Flourine needed?

Hint no of moles must be the same, tho you don't need to calculate that figure.

and for rest of problem, since PV=nRT where V and T are same, R is constant you have relation where P is directly proportional to n.

How do you know the number of moles will be the same?

How do you know the number of moles will be the same?

Because... that's the ideal gas law! :tongue2:
PV=nRT
R is a constant.

For a given pressure, volume, and number of moles of an ideal gas, the temperature can be calculated.

For a given pressure, temperature, and number of moles of an ideal gas, the volume can be calculated.

For a given volume, temperature, and number of moles of an ideal gas, the pressure can be calculated.

For a given pressure, temperature, and volume, the number of moles of an ideal gas can be calculated.

In other words, given 3 of the following 4 parameters: P,V,T,n, you can calculate the value of the 4th. Since in your problem, P,V, and presumably T are being kept constant, n must also stay the same.
Of course, it's a poorly worded problem, because it makes no mention of what happens to the temperature - leaving us to assume it remains the same. However, you could probably argue that you could use twice as many moles of another "ideal" gas, provided you have 1/2 the absolute temperature. (since it's "a tank", we can probably assume the volume of the tank remains the same.)

Correct method of solving.

PV=nRT
Since V,R, and T will be constant for both cases.

(P/n)_1=RT/V=(P/n)_2

So n_2 can be solved for, and then m_2 can be found.

yeah bpw91284 has it right. drpizza, the pressure isn't staying the same so you have to solve for a different number of moles. so yeah it's like I said: P_1/n_1=P_2/n_2

But it was broken down into 2 problems where the exchange of gasses would have required maintaining n at the same pressure. Please before taking others to task, read all the posts--if for no other reason than not to confuse the OP.

But it was broken down into 2 problems where the exchange of gasses would have required maintaining n at the same pressure. Please before taking others to task, read all the posts--if for no other reason than not to confuse the OP.

So you think that n_O2=n_F2?

Equal number of moles of gas occupies the same volume. As in at stp, one mole of any gas occupies a volume of 22.4dm3.

Equal number of moles of gas occupies the same volume. As in at stp, one mole of any gas occupies a volume of 22.4dm3.

PV=nRT
n=PV/RT

Same volume, different pressure, different amount of moles.