DeltaU = mCv(Tf-T1) Cv is in units of kJ/kg*K
Then for part b if i consider the case were the piston does not move, i have W = 0 (no change in volume) and QN2 = QHe (again Q cancels out) and I'm left with the exact equation i had before, and get the same answer: Tf = 57.2
Right. its a closed system, so it does not interact with the surrounding environment. so the volume of the system is constant (2 cubic meters). but each gas is allowed to expand or contract within the system.
but unless I am mistaken, there are still two unknowns and only one equation (final...
I don't have my book in front of me right now, but let me see if i can at least set this up right.
I have DeltaU = 0 (for a closed system)
DeltaU = (Q + W)N2 - (Q + W)He = 0
Q = m(Tf - T1)Cv
W = -P(Vf - V1)
doesn't this give me too many unknowns? Also, why am i using Cv? if the piston...
Thank you Kuruman, i figured it was some stupid mistake like that. That give me a much more reasonable answer of T2 = 54.78 degrees centigrade.
there are many forms of the ideal gas law, it depends on what units your R is in. Since this problem uses mass, i chose to use Individual Gas...
Homework Statement
Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow gas to leak into the other side. Initially, one side of the piston contains 1 cubic meter of Nitrogen (N2) at 500 kPa and 80 degrees...
Homework Statement
An electric motor consumes 12.0 KJ of electrical energy in 1.00 min. If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 3000 rpm...
ok, i corrected the moment of inertia, but i still didn't get the right answer... I took another look at linear momentum and figured this thing out.
I*w = m1v1r not where r = L/2
I*w = m1v1 like previously stated.
Thanks for your help. problem solved.
I calculated the velocity of the first ball to be v = (2gh)1/2 or 16.27
Then the moment of inertia for the system to be I = m1(L/2) + m2(L/2) + (1/12)mbarL2 or 57.12
Using these numbers i got w = 2.00 and v2 = 10.8 and found that the ball goes 1.48 m high... but this isn't the right answer.
there was no picture to go with the problem. I wrote word for word from my physics book. I assume the bar is horizontal and that ball travels strait up, but the problem does not state that.
Homework Statement
A 5.80 kg ball is dropped from a height of 13.5 m above one end of a uniform bar that pivots at its center. The bar has mass 8.50 kg and is 5.40 m in length. At the other end of the bar sits another 4.00 kg ball, unattached to the bar. The dropped ball sticks to the bar...
Oh.. my.. god. It worked. I never would have thought to set Va = Vb I though i was working 2 equations with 3 unknowns.
I got Va = Vb = 2.33 m/s
Elastic Potential = 105 J
Thank you!
ok i tried PE = 0.5mavai2 -0.5mavaf2-0.5mbvbf2 and got 3.55 J. That wasn't right either...
If the compression of the springs was at an maximum, wouldn't the relative velocities be zero? that was my intial guess and that wasn't right either.
Homework Statement
Blocks A (mass 3.50 kg) and B (mass 10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 9.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after...