Temperature & Gas: Correct or Incorrect?

[dorothy]

Homework Statement

Which of the following statements is/are correct when the absolute temperature of a gas is doubled?
(1) Average potential and kinetic energies of the molecules doubled.
(2) The gas volume will be doubled.
(3) The average velocity of the gas molecules will be increased but not doubled.
A. (1) only
B. (3) only
C. (1) and (2) only
D. (1) and (3) only

Relevant Equations

/

(1) Incorrect, since the root mean square speed(c^2）is directly proportional to the temperature, but it will just affect the average kinetic energy, doubling the k.e., but not related to the potential energy.

(2) Correct, since the volume of gas is directly proportional to the temperature

(3) Incorrect, since velocity is a vector, it is directional. Gas molecules move in different directions so the statement is not true.

These are my points but seems there is no such option. May I ask what is wrong with my attempt? Thank you.

 

[phinds]

(2) Correct, since the volume of gas is directly proportional to the temperature

I must have missed something. Where does it say that the volume is allowed to change?

 

[dorothy]

I must have missed something. Where does it say that the volume is allowed to change?

oh, so I should assume the volume is fixed if the question doesn’t mention about it. Is that right?

May I also ask about (1) & (3) as well? thanks

 

[phinds]

oh, so I should assume the volume is fixed if the question doesn’t mention about it. Is that right?

That's certainly what I would do.

 

[Steve4Physics]

IMO the question is unclear/badly written. Have you given us the entire question, word-for-word?

Homework Statement:: Which of the following statements is/are correct when the absolute temperature of a gas is doubled?

Some answers will depend on whether the gas is ideal or real. Do you know which it is supposed to be?

(1) Average potential and kinetic energies of the molecules doubled.

The question won't apply to an ideal gas which has no potential energy.

(2) The gas volume will be doubled.

That depends on
a) if the pressure is kep constant
b) whether the gas is ideal or real

(3) The average velocity of the gas molecules will be increased but not doubled.

I think this is a 'trick' question. What is the average of all of the velocities of the particles in a (stationary) container of gas?

Or it could be badly written and actually mean 'average speed'.

(1) Incorrect, since the root mean square speed(c^2）is directly proportional to the temperature, but it will just affect the average kinetic energy, doubling the k.e., but not related to the potential energy.

The answer depends on whether the gas is ideal or real.

(2) Correct, since the volume of gas is directly proportional to the temperature

Only if the pressure is constant and the gas is ideal.

(3) Incorrect, since velocity is a vector, it is directional. Gas molecules move in different directions so the statement is not true.1

Correct - but I think you have not fully picked-up the key point (see above)! Or ir may be a badly written question and is really about average spped.

Have you learned about both ideal and real gases? If not, you may be expected to answer the question on the assumption it is about only an ideal gas.

 

[haruspex]

so I should assume the volume is fixed

No, you don’t need to assume that. The question asks what you can deduce from the given fact. Since you are not told anything about the volume or pressure, before or after, you cannot deduce the volume doubles.

The question won't apply to an ideal gas which has no potential energy.

Including potential energy doesn't stop it being a valid statement for an ideal gas. Rather, it implies you should not assume an ideal gas.

Or ir may be a badly written question and is really about average spped.

Unfortunately, that seems to me the most likely.

 

[vela]

(1) Incorrect, since the root mean square speed(c^2）is directly proportional to the temperature, but it will just affect the average kinetic energy, doubling the k.e., but not related to the potential energy.

Have you discussed monatomic gasses, diatomic gasses, and the equipartition theorem yet?

I should assume the volume is fixed if the question doesn’t mention about it. Is that right?

No, you generally shouldn't assume anything about the volume. The question is asking: if you double the temperature, will the volume always double? As others have pointed out, the volume doubles only under certain conditions, so you can't conclude that doubling the temperature necessarily results in the volume doubling.

The question won't apply to an ideal gas which has no potential energy.

If you double zero, it's still zero.

 

[Steve4Physics]

If you double zero, it's still zero.

Yes indeed. And in fact I considered that when writing the post.

However I felt uncomfortable doubling something that was necessarily zero, so decided to phrase it the way I did.

 

[Redbelly98]

If you double zero, it's still zero.

I'm not aware of how the potential energy of an ideal gas is even defined. Am I missing something?

(OP has already ruled out this statement as incorrect, so I think it's fair game for discussion.)

 

[haruspex]

I'm not aware of how the potential energy of an ideal gas is even defined.

Same as its definition for a non-ideal gas, but with the added fact that it is constant?

 

[Redbelly98]

Same as its definition for a non-deal gas, but with the added fact that it is constant?

Okay, fair enough.

 

[vela]

I'm not aware of how the potential energy of an ideal gas is even defined. Am I missing something?

(OP has already ruled out this statement as incorrect, so I think it's fair game for discussion.)

One of the assumptions of an ideal gas is that the particles don't interact, so I'd just take the PE as zero.

 

[Redbelly98]

One of the assumptions of an ideal gas is that the particles don't interact, so I'd just take the PE as zero.

Yeah, I was initially thinking -- erroneously -- it would be undefined for noninteracting objects. But okay, if the forces between the particles are zero then it's just a constant. And then apply the usual convention of calling it zero in the limit of infinite separation.