Recent content by chocopanda

Hi yes, that's what I got as well as per my previous post for the partial pressures, and I have to add them up for the overall final pressure

Thank you so much for the quick answer! I figured out where I went wrong (I think): Instead of calculating the final (partial) pressure via the final temperature, I mistakingly used the gases inital temp. and volumes. So to calculate the partial pressures, which should be equal like you...

Hey everyone, I have an attempt at fully solving this problem (my final pressure is ##p_f = 5373,64 hPa##, final temp. is ##T_f = 303,15K = 30C##), but this exercise confuses me very much. First, I have not used the masses in my calculations and I'm pretty sure my prof. accidentally copypasted...

Thank you, so the equation would be $$T_f = \frac {V_1*T_1+V_2*T_1} {V_1+V_2} $$ Hope there aren't any mistakes in it And for the final volume ##V_f##, since the air cools down to around ##T_f=45,5C##, I can't just sum up ##V_1+V_2## since the overall volume decreases as well

I completely overlooked that we have different ##n := n_1, n_2## I'm so sorry. I feel like I'm getting very close to ##T_f## :) Editing the equation: $$5/2 * n_1*R*(T_f - T_1) = 5/2 * n_2*R*(T_2 - T_f)$$ Simplifying and solving for ##T_f##: $$n_1*(T_f - T_1) = n_2*(T_2 - T_f)$$ $$T_f = \frac...

Since the ideal gas law ##p*V=n*R*T## applies. ##n## is the amount of particles in a gas in moles. Each gas has a certain amount of moles ##n_1## and ##n_2##, solving for ##n##: ##n_i= \frac {p*V_i} {R*T_i}##

Thanks for the hints, I'll try again. So the internal energy of a system is: ##U = f/2 * n*R*T## and since air is not an monoatomic gas we can insert f=5? The change of the internal energy ##\delta U_1 = -\delta U_2## from the hotter gas to the colder and since a change in ##U## means ##\delta...

I'm very sorry, I really do not know where to go from here... is this possible with the ideal gas law? I know that both gases have the same pressure. I could solve for the pressure p for both gases: p1=n1*R*T1/V1, p2=n2*R*T2/V2 The mixing process happens under a constant pressure p which should...

Thanks for replying :) I thought about that, but the exercise requires that both gases should be mixed and calculate the final volume at the end, I'm still struggling :(

Hey, thanks for replying! Yes, we have learned about the ideal gas law. I am having trouble applying it. For this case, I would perhaps choose the form p*V = n*R*T to get the amount of gas in moles for both gases and maybe sum them up? As in n1 + n2 = n (sum of all the gas in moles, since the...

To be honest, thermodynamics is really not my strong suit and I get confused when and how to apply formulas. My thought process is as follows: - there are two ideal gases (ideal gas law applies) - the pressure remains constant (isobaric process), so p1= p2 = p - I imagine there being two...

Hello DrClaude, thank you for replying. I tried to do what you suggested: $$| \langle l|p|n \rangle |^2 = \langle n|p^2|n\rangle = \frac{h}{2mw} (2n+1) $$ That's my result. How would I continue?

I'm trying verify the proof of the sum rule for the one-dimensional harmonic oscillator: $$\sum_l^\infty (E_l-E_n)\ | \langle l \ |p| \ n \rangle |^2 = \frac {mh^2w^2}{2} $$ The exercise explicitly says to use laddle operators and to express $p$ with $$b=\sqrt{\frac {mw}{2 \hbar}}-\frac...

I'm wondering if I calculated that correct and if I can still simplify it because I don't know how that would work :)

Hello everyone, I'm new here and I'm struggling with the mathematical formalities in quantum mechanics. $$\langle n+1|b^\dagger bb^\dagger + \frac 12 |n \rangle = \langle n+1|b^\dagger bb^\dagger |n \rangle + \langle n+1| \frac 12 |n \rangle $$ $$ = \langle n+1|b^\dagger b \sqrt{n+1} |n+1...