Finding the Remaining Pressure of a Gas After Removing 2/3rds

[Const@ntine]

Homework Statement

In a container we have a gas, at P = 11.0 atm & T = 20.0 C. We remove 2/3rds of the gas and increase the temperature to T = 75.0 C.What's the pressure of the gas that remains?

Homework Equations

PV = nRT

The Attempt at a Solution

Normally this would be a regular "put the numbers in the formula" exercise, but I'm having trouble understanding just what "remove 2/3rds of the gas" means. If the gas was, say, a block of wood, I'd be removing 2/3ds of its mass. With that logic, I figured it was the n (moles) that were decreased, and that the volume V was kept the same (same container). But I didn't get the wanted result from this, so I was wrong.

Then I figured maybe the change happens so fast that the gas doesn't manage to spread and have the same Volume again, so both n & V should be n' = n/3 & V' = V/3. That didn't work either. So I'm kinda stuck on what that phrase means.

Any help is appreciated!

 

[Daniel Gallimore]

I think you're original method is on the right track, but you didn't include your final answer in your post, so I'm going to have to guess that what I did here is different.

Pretty much baked into the definition of the gas is the property that it expands to fill any container, so I wouldn't worry about changing the volume of the gas. Besides what if I took molecules randomly out of the gas instead of scooping out the top $2/3$? That's also a viable interpretation of "remove $2/3$."

The first thing we need to do is solve for $n_iR/V$ in terms of the initial quantities. Doing so gives $P_i/T_i$. Now we need to solve for $n_iR/V$ in terms of the final quantities, where $n_f=n_i/3$: $$P_fV=n_fRT_f=\frac{n_iRT_f}{3}$$ so $n_iR/V=3P_f/T_f$ Now we equate the two representations of $n_iR/V$ and solve for $P_f$: $$\frac{P_i}{T_i}=\frac{3P_f}{T_f}$$ so $P_f=P_iT_f/3T_i$. At this point, feel free to plug in the numerical quantities and count your sig figs. Right now it looks like if you plugged everything in as is, you'd get a pressure in $\text{atm}$, so no need to change units unless you really want to.

 

[Const@ntine]

I think you're original method is on the right track, but you didn't include your final answer in your post, so I'm going to have to guess that what I did here is different.

Pretty much baked into the definition of the gas is the property that it expands to fill any container, so I wouldn't worry about changing the volume of the gas. Besides what if I took molecules randomly out of the gas instead of scooping out the top $2/3$? That's also a viable interpretation of "remove $2/3$."

The first thing we need to do is solve for $n_iR/V$ in terms of the initial quantities. Doing so gives $P_i/T_i$. Now we need to solve for $n_iR/V$ in terms of the final quantities, where $n_f=n_i/3$: $$P_fV=n_fRT_f=\frac{n_iRT_f}{3}$$ so $n_iR/V=3P_f/T_f$ Now we equate the two representations of $n_iR/V$ and solve for $P_f$: $$\frac{P_i}{T_i}=\frac{3P_f}{T_f}$$ so $P_f=P_iT_f/3T_i$. At this point, feel free to plug in the numerical quantities and count your sig figs. Right now it looks like if you plugged everything in as is, you'd get a pressure in $\text{atm}$, so no need to change units unless you really want to.

That's exactly what I did, but I missed the fact that I had to convert the temperatures from Celcius to Kelvin, so my result didn't match with the book's. That said:

P_i = 11.0 atm
T_f = 75.0 + 273.15 = 348.15 K
T_i = 25.0 + 273.15 = 298.15 K

P_f = P_i*T_f/*T_i = 11.0atm * 348.15 K/3*298.15 K = 4.28 atm

Thanks a ton for the help!