Ideal Gas Situation: What happens when P, V decrease?

AI Thread Summary
When the volume and pressure of an ideal gas decrease, the temperature also decreases, leading to a reduction in internal energy. The internal energy is directly related to kinetic energy, which decreases as temperature drops. Intermolecular forces remain unchanged and are effectively zero in an ideal gas scenario. The initial confusion about the relationships between internal energy, temperature, and kinetic energy was clarified, confirming that a decrease in temperature results in a decrease in internal energy. The final understanding aligns with the principles governing ideal gases, resolving the initial errors in reasoning.
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Homework Statement



If the volume V and pressure P of an ideal gas decrease, which statement is true about the internal energy IE, temperature T and the intermolecular forces IF?

(I am not going to copy and paste the choices here in an effort to focus attention on the attempt of solution.) Would greatly appreciate if anyone can point out an error in thinking!

Homework Equations



Change in Internal Energy (I.E.) = Q (heat added) + W (done on system)
PV/T = PV/T

The Attempt at a Solution



First: My final answer. (Incorrect)
IE increases, T increases, IF are unchanged

Since P and V decrease, by P1V1/T1 = P2V2/T2, the temperature of this ideal gas should decrease.
Since Kinetic Energy is directly proportional to KE, the Internal Energy of the system should increase!
(In an ideal gas: no Intermolecular forces, otherwise, the gas would liquefy!)
So, Internal Energy is composed of KE, and since Temperature goes up, IE goes up, and IF is constant (or zero).

Many thanks in advance !
 
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Welcome to PF, yaylee! :smile:


yaylee said:

Homework Statement



If the volume V and pressure P of an ideal gas decrease, which statement is true about the internal energy IE, temperature T and the intermolecular forces IF?

(I am not going to copy and paste the choices here in an effort to focus attention on the attempt of solution.) Would greatly appreciate if anyone can point out an error in thinking!

Homework Equations



Change in Internal Energy (I.E.) = Q (heat added) + W (done on system)
PV/T = PV/T

The Attempt at a Solution



First: My final answer. (Incorrect)
IE increases, T increases, IF are unchanged

Since P and V decrease, by P1V1/T1 = P2V2/T2, the temperature of this ideal gas should decrease.

That is correct: the temperature decreases.

Since Kinetic Energy is directly proportional to KE, the Internal Energy of the system should increase!

Huh? I'm assuming the with KE you mean Kinetic Energy, don't you?
This doesn't make sense to me.

Either way, the Kinetic Energy is directory proportional to the Internal Energy.
And they are both also directly proportional to the temperature.
So since temperature goes down, so do KE and IE.

(In an ideal gas: no Intermolecular forces, otherwise, the gas would liquefy!)

It's not true that the gas would liquify - that would depend on the circumstances.
But yes, in an ideal gas there are no intermolecular forces.

So, Internal Energy is composed of KE, and since Temperature goes up, IE goes up, and IF is constant (or zero).

Many thanks in advance !

Hold on!
Didn't you say before that the temperature decreased? :confused:

Btw, you are right that IF is constant, moreover it is and remains zero, since we're talking about an ideal gas.
 
Hello I like Serena !

Thank you for the welcome.

I am sorry for mixing up increase/decrease throughout the problem.

In an ideal gas, As temperature DECREASES, Kinetic Energy decreases and therefore internal energy DECREASES.

Also: Intermolecular forces are unchanged, or remain zero throughout.

I am wondering why this is getting marked incorrect, oh well! Thank you once again.
 
Ah, OK. There was a mistake with the scoring. Your answer was correct. Thanks for the help!
 
As you just wrote it down, it is correct.

Cheers!
 
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