Recent content by lesodk

yes sorry, that was what i meant. Do you know how to show the formula from the equations of motion?

I have this formula in my book: \[ a = - \omega x^{2} \] Can anyone help me show how to show that this formula is true?

my point is just that after v3 and p3, you go back to the initial place v1 and p1. Thats what i mean by 3-1.

thanks a lot

Nope, i don't have that information.

yes i have

so i shoud insert the value for dQ = nCvdT and by that getting dS = int( (n Cv dT)/T) = n Cv ( (1/T) dT ) = nCv ln(T2/T1) ? Is that correct, even though the process is not reversible?

entropy change = dS = int( dQ/T ) where you integrate from state 1, to state 2. The forumula only holds for reversible processes and i don't know if that's the case in my question.

There is nothing there about entropy change

How do i calculate the entropy change for an isobaric process, if i know T1, T2, n and Cv ? Thanks,

Homework Statement A Cyclic proces of three parts for a monoatomic idealgas: 1-2: Isochor, where: p2 = 2*p1 v2 = v1 2-3: Adiabatic, from v2 = v3, where: v3 > v2 p3 = p1 3-1: unknow proces, where v4 = v1 p4 = p1 My problem is to determine Q1, Q2, Q3 and W1, W2, W3 and...

I get: c) dU = -W = -1.25*10^4*kJ does this seem to be a likely result?

i use the following for my calculations: a) K_tr1 =3/2 nRT = 3/2 pV =3/2*1,013*10^5 Pa * 0,005m^3 = 759,75J b) nRT=pV <=> n = pV/RT = (1,013*10^5 Pa * 0,005m^3)/(8,314 J/mol*K * 300K) = 0,2 mol m = n*M = 0,2*2,016g/mol = 4*10^-4 kg K_tr_2 =1/2mv² = 1/2 *4*10^-4kg*(300m/s)² = 18,4JTotal...

i meant translational kinetic energy.

I need to solve this problem. A big balloon with volume V1 = 2.00 * 10^3 m^3 contains helium at 15 degrees celsius and 1 atm. The balloon now rises from the ground to a height 'h' where the pressure is 0.900 atm. The process is adiabatic: a) find the volume at height h b) find the...