<p>PE doesn’t matter for the actual graph, but is a key part in understanding KE. I just know that isn’t not a concave down parabola. That’s not what I put. My end pints were back on the x axis because…well…the velocity is at zero.</p>
<p>oh, yeah exactly, it is upside down parabola ending at 0.5 and -.5.</p>
<p>Oh, wait…</p>
<p>what did you guys say about the volume and pressure?
I said volume increases because the work is done on the system.
pressure stays the same
U=3000(heat though i am not sure if it is the correct number) + 1230 =4230?</p>
<p>I put volume decreases.
W=-PV Work is positive therefore the final volume must be smaller than the original volume.</p>
<p>When work is done on a gas the gas is compressed.</p>
<p>i meant volume decreases, sorry about that</p>
<p>what about pressure? pressure stays the same right?</p>
<p>I wrote that pressure stayed the same. </p>
<p>Although, it might be that pressure decreases because the temperature increased due to the added energy.</p>
<p>screw ap physics. I am pretty sure i got a 5.
In the end, you have to get 50 mcq questions correct and 50/80 on free response to score a five.</p>
<p>Pressure has to increase bro. higher temperature+ lower volume is a recipe for more frequent and more forceful collisions (from a chemistry standpoint)</p>
<p>Did anyone get the mass of the block in that spring question to be 4.22 kg?</p>
<p>@ds I meant increases. PV=T</p>
<p>@ds Yes, I got ~4.1kg. Difference is most likely due to rounding in previous steps.</p>
<p>Was the change in internal energy for when work was done on the gas, but the temperature stayed constant, 0? Looked a bit awkward in that giant answer space.</p>
<p>And yes, pressure should increase, although I was looking for “an ideal gas is placed in a container”…</p>
<p>I got 4.1 as well.</p>
<p>No I got that it was equal to the work done.
U=Q + W</p>
<p>Temperature stayed the same so I assumed Q=O
U=W</p>
<p>But…what do I know at this point.</p>
<p>Yeah, the large space provided was a bit deceiving, but the change in internal energy was indeed zero. The only information relevant for that part was that temperature is constant.</p>
<p>I also put down the work done for U. The temperature remained constant so I put 0 for Q.
So U = W, though i guess that’s wrong</p>
<p>U is a property of the gas itself; temperature is also a property of the gas (AvgKE), which remains constant. Change in U of the gas depends solely on the change in temperature of the gas, which in this case is zero. Q them equals -W, therefore, heat has been removed.</p>
<p>Glad I’m not the only one :)</p>
<p>what was the pressure question out of?</p>