2-74
This is almost exactly like the one I did in class. First find the volume of Tank B and the mass of Tank A using the ideal gas law. VB = (mRT1/P1)B = (5 kg)(0.287 kPa-m3/kg-K)(308 K) = 2.21 m3 mA = (P1V/RT1)A = (500 kPa)(1.0 m3)/(0.287 kPa-m3/kg-K)(298 K) = 5.846 kg Final volume = VA + VB = 1.0 + 2.21 = 3.21 m3; total mass = mA + mB = 5.846 + 5.0 = 10.846 kg. So Pfinal = mRTfinal/V = (10.846 kg)(0.287 kPa-m3/kg-K)(293 K)/3.21 m3 = 284.1 kPa.

2-102
You need to find the initial mass, final mass and the difference between the two.
M1 = P1V/RT1 = (800 kPa)(20 m3)/(0.2968 kPa-m3/kg-K)(298 K) = 180.9 kg M2 = P2V/RT2 = (600 kPa)(20 m3)/(0.2968 kPa-m3/kg-K)(293 K) = 138.0 kg The change in mass (or the amount escaped) is 42.9 kg.

3-20
This is a matter of finding the rate of heat transfer from convection and radiation and adding them together. Assume that the Q's have dots over them. Qconv = hA(Ts - Tf) = (15 W/m2-°C)(0.007854 m2)(70-20 °C) = 5.89 W. Qrad = esA(Ts - Tsur) = 0.8(5.67 X 10-8 W/m2-K4)(0.007854 m2)*((343 K)4 - (293 K)4) = 2.31 W. The total rate of heat transfer is 8.20 W. Note, to find the surface area of the sphere, use 4pr2.

3-28
Use the conduction heat transfer equation: kT = (Q * Dx)/(A * DT). 500 w/m2 is Q/A, so kT = (500 W/m2)(0.02 m)/(100 °C) = 0.1 W/m-°C.

3.38
You are given the initial and final volumes. Use the function for P to find the initial and final pressures.
P1 = (-1200 kPa/m3)(0.42 m3) + 600 kPa = 96 kPa. P2 = (-1200 kPa/m3)(0.12 m3) + 600 kPa = 456 kPa. If you find the area under the curve, just draw it and you get -82.8 kJ. If you integrate, W = ò (aV + b) = 1/2a(V22 - V12) + b(V2 - V1) = -82.8 kJ. Note the negative sign. The surroundings is doing work on the system so the sign is negative.

3-45
a) The final pressure is found from:
P3 = P2 + F/A = P2 + kx/A. Putting in numbers: P3 = 150 kPa + (100kN/m)(0.2m)/0.1 m2 (1 kPa/1kN/m2) = 350 kPa. b) To find the final temperature, you need two independent, intensive properties to set the system. You have pressure - your other one is specific volume. Your final volume is V2 + xA = 0.2 m3 + (0.2m)(0.1 m2) = 0.22 m3. The specific volume is V/m = 0.22 m3/50 kg = 0.0044 m3/kg. Looking up the pressure at 350 kPa, you find that vf = 0.0010 m3/kg and vg = 0.5243 m3/kg. Since your specific volume is in between, you have a saturated mixture, and your T = Tsat @ 350 kPa = 138.88 °C. What if your specific volume has been less than 0.0010. Then you would have a compressed liquid. If it was above 0.5243, you would have a superheated vapor. c) To find the work, just draw the curve. You have constant pressure from the initial to the intermediate stage, so you have a rectangle. Then the pressure changes linearly, so you have a triangle. Add up the areas of the geometric figures, and you should get a work of 27.5 kJ. Again, note the positive value. The work is done by the system.