Recent content by chocolatepie

Thank you so much for your help! I looked through my notes, and I realized that I actually had that formula LOL. Anyway, thank you again! Take care! :smile:

Whoa! I got: n= 0.02 for both initial and final P2 = 62 kPa ΔS = 0.12 J/K However, I did not learn that formula.. I've seen the first and second part of the equation separately but not together like that. Shouldn't I use the second part of the equation for the reversible isothermal...

Oh OK. I see what you mean now :redface: But when determining the number of mole using PV=nRT, how do I know which T or which V I need to use? I have 2 values (for example, V1 and V2, etc). This is actually why I used the change (Δ).

Uh.. I wrote 3/2R because I was using the equipartition theorem. Since the monoatomic ideal gas does not have neither rotational nor transitional energy, 3/2RT. Then I sub this into heat capacity formula where T would get cancelled. I am not even sure if I am doing this question right...

Thank you for start replying my question. I REALLY appreciate it! Ok. what I would do is to assume that pressure is constant, so I would use the formula I posted earlier. C p,m = 3/2R C p = 3/2R x change in mole (determined by PΔV=ΔnRΔT) which would give me 0.9989 J/K Then I would find the...

Homework Statement A monoatomic ideal gas is initially trapped in a 0.500L container at 298K and 100kPa. The sample is then allowed to expand to 1.00L and the temperature is increased to 373K. Determine change of the entropy (ΔS) Homework Equations ΔS = ∫(T1 to T2) dqrev/T The Attempt...

Homework Statement none Homework Equations none The Attempt at a Solution maybe they are broken down just like glucose in metabolism?

Now I have: (a) qv = 3325.6J (b) qp = 3525.79J ?!

Awesome! thank you so much! Now I only need to deal with the other question.. :)

At constant V, I had a formula ΔU = qv Does it mean I need to add ΔU for each gas since they are in a mixture? Also, I am using ΔU +PΔV =q for the part (b). There is no work value given.. and so is pressure

Could you explain why it is 5?

To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?

Doesn't oxygen have 6 degrees of freedom and Ne has 3?

Could you check my math in my previous post?

Okay, that makes sense. Now I have the following: (a) -1993J/mol (b) -23.3 x 10^3 J/mol (c) -25293J/mol (d) -23.3 x 10^3 J/mol yes? :)