1. Pressure on lungs underwater
-Is the height at the surface 0? I guess that is a bit relative to how you approach the question.
-What is the pressure at height 0 (or the surface)? Is it 1 atm??
@& Right. You get to put y = 0 at any point you wish. The change in y from the surface of the water to the lungs is the same, no matter what reference point you use.
For example of the bottom of the lake is 10 m below the surface, with the lungs 3 m below, then you could use the bottom of the lake as your y = 0 point. y_surf and y_lungs would be 10 m and 7 m, respectively. rho g y_lungs - rho g y_surface would be rho g * 7 m - rho g * 10 m = -rho g * 3 m.
Or you could use the surface of the lake as your y = 0 point. y_surf and y_lungs would be 0 m and -3 m, respectively. rho g y_lungs - rho g y_surface would be rho g * (-3 m) - rho g * 0 m = -rho g * 3 m, the same as before.
Similarly you could choose y = 0 at the position of the lungs. You would still get the same result for rho g y_lungs - rho g y_surface.
*@
2. I ended up with an answer of 940.9K or 667.91 C
@& You need to include a brief synopsis of the question. Tests are randomly generated and I have no way of knowing what the conditions of the problem were.*@
3. T_1 = 562.26K
T_2 = 711.72 K
'dT [0 -> 1] = 264.26K
'dT [1 -> 2] = 149.46K
I can get the temperature and bernoulli's equation but I am not understanding the heat change. I got 'dQ [0 -> 1] = 660.65*nR and 'dQ [1 ->2] = 373.65*nR for my change in heat. How many moles were there of the gas?
@& I need more information on what information was given in the question.
The number of moles of gas can be found if you know P, V and T. n = P V / (R T).
Specific heat is 3/2 R for monatomic gas at constant volume, 5/2 R if diatomic. If pressure is constant then it's 5/2 R for monatomic and 7/2 R for diatomic.
That appears to be consistent with the numbers you give for the first part of the cycle, assuming the gas to be diatomic. 5/2 n R * (264 C) = 660 Celsius * n R.
I suspect you used 5/2 n R T for the second part of the cycle; that part was most likely at constant pressure and you should have used 7/2 R.*@
4. My answer for P_1 was 78760 kPa. What is y_2? Not sure how that is found.
@& I can't conjecture anything abou the problem from this information.*@
5. The grams per particle of each gas is found by dividing the mass by Avagadro's number (6.023x10^23). This gives us g/particle. What is ""k""?
@& P V = n R T
P V = N k T
n is the number of moles, N is the number of particles.
It follows that k = R / Avagodro's Number.*@
6. Do you have any references to the problem sets or class notes that relate to this problem?
@& I need to know what the problem asks.*@
7. It's been a while since I had calculus but i'll take a stab at this one.
'd'rho g h + 'd 0.5 'rho v^2 + 'dP = 0
P' = 'rho^2 g h + 'rho^2 0.5 v^2
I'm not really sure how to approach this.
Thank you for any help and clarification.
@& For a function f(x), df = f ' (x) dx.
This denotes the change expected in the value of f, given a change dx in the value of x, in the vicinity of the point x.
f ' (x) is the rate of change of f with respect to x, dx is the change in x, so f ' (x) dx is the approximate change in the value of f. The reason it's approximate is that f ' can't be expected to stay constant while x changes by `dx; but if `dx is small, you don't expect a big change in f ' (x).
It would follow that d( f * g) = (f * g) ' dx. By the product rule this would be ( f ' * g + f * g ' ) dx. This could be written g * f ' (x) dx + f * g ' (x) dx, and since f ' (x) dx = df, whiole g ' (x) dx = dg, we can say that
d(f g) = g df + f dg.
Extending this to the expression rho g h we could write
d(rho g h) = rho g dh + rho h dg + g h d rho.
The change in rho g h is the sum of rho g times the change in h, rho h times the change in g and g h times the change in rho.
This all follows from the product rule.
Now in most applications for your course, we consider density rho and acceleration of gravity g to be constant, with no change, so d g = 0 and d rho = 0.
So
d (rho g h) = rho g dh + 0 + 0, or just rho g dh.
d(.5 rho v^2) = .5 d(rho v^2) = .5 ( v^2 d rho) + rho d(v^2) ).
d rho is typically zero.
d(v^2) = 2 v dv, which can be understood by the power rule for the derivative of by considering v^2 to be v * v, and applying the product rule.
So if rho is constant, d(.5 rho v^2) = .5 * (2 v dv) = v dv.
dP is just dP.
The differential is therefore
rho g dh + v dv + dP = 0.
This is useful if you have small changes in v, h and P.
Note that rho is pretty much constant for fluids; if it varies it's mostly due to temperature changes.
rho isn't constant in most situations involving gases. It varies with pressure and temperature. However we don't deal with such systems in this course.
g is pretty much constant unless you're talking about big changes in altitude; if you move further from the center of the Earth the acceleration of gravity does diminish, and in such situations you might need to consider rho h dg.
*@
@& I've inserted the best answers I could with the given information.
On a couple of problems I really need more information about what the problem gives you and what it's asking for. I typically also want to see at least an outline of your reasoning, as opposed to just a final answer.
I'll be glad to answer additional questions, including questions about the problems I couldn't answer here; just be sure to include sufficient information.*@