EK Physics 1001, Question 525

[sillyjoe]

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525. A rigid container holds air (density 1.3 kg/m ) at 0 °C. If the temperature is increased to 273 °C, what is the new density of the air?

A. 0.65 kg/m^3
B. 1.3 kg/m^3
C. 2.6 kg/m^3
D. 3.9 kg/m^3

B is correct. The volume and mass remained the same, so the density remains the same. p = m/V

My understanding from TBR is that the real volume of a gas is not really the entire container but rather the subtraction of the container - free space. Is this question assuming that the volume of the container will always equal the volume of gas?

 

[t5Nitro]

Your understanding is correct for a real gas. Unless you are told otherwise, you can assume you are dealing with an ideal gas where the atom volume is not taken into account, and the volume is considered to be the volume of the container.

From the question you were given, and the ideal gas law equation, you know temperature increased (so the right side of the equation increases). On the left side of the equation, since it is a rigid container, volume remains the same so pressure must increase to balance the right side increase.

Overall, you just know that temperature and pressure increases. No changes have been made to mass or volume which is what you just used to solve the problem.

 

[sillyjoe]

Your understanding is correct for a real gas. Unless you are told otherwise, you can assume you are dealing with an ideal gas where the atom volume is not taken into account, and the volume is considered to be the volume of the container.

So I just overthought this? Thank you for your help.

 

[t5Nitro]

So I just overthought this? Thank you for your help.

No problem, I added more to my first post (I seem to do this a lot lately - haha). Let me know if that helps clarify better.

 

[sillyjoe]

No problem, I added more to my first post (I seem to do this a lot lately - haha). Let me know if that helps clarify better.

Just help me clarify, how does an increase in pressure then not change the density?

 

[t5Nitro]

Just help me clarify, how does an increase in pressure then not change the density?

Yeah, seems like it would. It's been a little while since I've studied for the MCAT, and some of the stuff isn't as sharp as it was when I was studying. But I just used the PV = nRT equation and knew 'n' and 'R' are constant. When T increases, P has to increase since 'V' is held constant by a rigid container. The density formula, p = m/V remains the same since the changing variables are not part of that. Or you could substitute V = nRT/P into p = m/V and get p = m/(nRT/P), but the proportional increase in T and P makes that ratio stay the same. I just used equations on this one, but if someone can put in a conceptual understanding, that would be easier to understand.

Ultimately, you have the same mass of gas (as long as it was a sealed container where some did not escape) within the same volume (since it is rigid). So it makes sense that density shouldn't change.