Recent content by CurtisB

The question asked for gauge pressure, which is the pressure at 2 minus atmospheric pressure given by p_2-p_3 in the top equation, which is 6.97*10^4. So yeah, the absolute pressure at 2 would be the gauge pressure plus atmospheric which is pretty much what you got.

Oh, thank you, I accidentally put in the volume flow rate in as v instead of the velocity of the fluid, I have the correct answer now, Thanks.

Homework Statement Water flows steadily from an open tank. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m^2; at point 3 it is 0.0160 m^2. The area of the tank is very large compared with the cross-sectional...

if pV=Nk_BT then \frac{p}{T} = \frac{Nk_B}{V} = constant so \frac{p_f}{T_f}=\frac{p_i}{T_i}, convert ^oC to K so \frac{p_f}{393.15}=\frac{p_i}{293.15} and because p=\frac{F}{A} and p_i = p_a, \frac{F}{A}=\frac{393.15p_a}{293.15} so F=\frac{393.15AP_a}{293.15}=1.34Ap_a and since the force on the...

Thanks everyone, that was really helpful, one quick question though, should Liters always be converted to meters cubed when Pascals are used for the pressure, and Liters used when the pressure is in atm?

Well the answer was actually F = Ap_a - (293.15Ap_a)/(393.15), can anyone tell me why the two forces are swapped around, int the first part it was the inside force minus the outside force so why is this the outside force minus the inside force, wouldn't the second part be a negative value...

Thanks, that really helped, I got the answer to part one to be 0.34Ap_a but now the second part is getting me. I work out the final pressure inside to be (293.15p_a)/(393.15) so the force out = (293.15Ap_a)/(393.15), then the force in equals -Ap_a so the final answer should be...

So for part one, if delta_U = 3/2(pV) then delta_U should equal 3/2* the change in p * V = 3/2(Ap_1 - p_1)V_1. does this look right. for part 2, W = pV*ln(V_f/V_i) then W = p_1V_1*ln(AV_1/V_1), or should p and V be the changes in p and V, the feedback I get when I tried that answer was that...

Homework Statement A monatomic ideal gas has pressure p_1 and temperature T_1. It is contained in a cylinder of volume V_1 with a movable piston, so that it can do work on the outside world. Consider the following three-step transformation of the gas: 1. The gas is heated at constant...

I tried n= (pV)/(RT) = [(7.6*10^4)*(2.7)] / [(8.31)*(297)] = 83.14 moles, that means that there is 83.14*30.1 = 2502.57g of gas left in the flask, that just doesn't seem right.

Homework Statement A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. 1-If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F_120 on the...

Homework Statement A flask with a volume of 2.70 L, provided with a stopcock, contains ethane gas (C_2 H_6) at a temperature of 297 K and atmospheric pressure 1.01×10^5 Pa. The molar mass of ethane is 30.1 g/mol. The system is warmed to a temperature of 396 K, with the stopcock open to the...

Yeah, OK. The correct answer is 1.14 e 4, that's no where near any of my calculations, what am I missing here?

Sorry, I actually ment to write 579617.6619, I was looking at the wrong part of my working out. If the sphere radiates that much energy won't that amount ov energy be required to be put back into maintain the sphere at that temperature? EDIT- sorry I completely missed the homework section of...

I forgot to mention that you have to take the Stefan-Boltzmann constant to be 5.67×10−8