Here are the solutions to the packet I handed out on Friday:
http://groups.google.com/group/bronx-ap-physics/web/Thermo+Review+Problem+Solutions.pdf
As requested, I will have a couple problems for you to play around with, but I expect most of what we do will come from you!
Sunday, December 21, 2008
Saturday, December 20, 2008
Bronx Tournament is ON!
The Bronx FLL Tournament is going on as scheduled today, Saturday, December 20, 2008.
Please leave yourself extra time to get to the event safely.
See you all there!
Please leave yourself extra time to get to the event safely.
See you all there!
Tuesday, December 16, 2008
Thermo-cycling in the Snow!
Handout answers:
1
a) T = 481 K
b) Wnet = +4000 J
c) For a full cycle, delta_U = 0. This means that for the entire cycle, Q = -W = -4000 J.
d) For any cycle, delta_U is zero, so it stays the same.
2.
a) AB,CD - isobaric, BC, DA - isochoric
b) 1.202 x 10^-4 moles (use P,V, and T at state D.)
c) Wab = -1.0 J, Wad = 0.5 J, all others zero
d) net work for cycle is the area inside = -0.5 J
e) Tb = 800 K, so delta_U = 0.750 J
f) P = Wnet/delta_t = 0.5 J/1.2 s = .417 W
3. b, c, a (remember that in c and a, the work is negative, and C has a work that is LESS negative.)
1
a) T = 481 K
b) Wnet = +4000 J
c) For a full cycle, delta_U = 0. This means that for the entire cycle, Q = -W = -4000 J.
d) For any cycle, delta_U is zero, so it stays the same.
2.
a) AB,CD - isobaric, BC, DA - isochoric
b) 1.202 x 10^-4 moles (use P,V, and T at state D.)
c) Wab = -1.0 J, Wad = 0.5 J, all others zero
d) net work for cycle is the area inside = -0.5 J
e) Tb = 800 K, so delta_U = 0.750 J
f) P = Wnet/delta_t = 0.5 J/1.2 s = .417 W
3. b, c, a (remember that in c and a, the work is negative, and C has a work that is LESS negative.)
Monday, December 15, 2008
In This House, we obey the laws of Thermodynamics!
Handout solutions:
1.
a) Tf = 400 K
b) The ratio of average kinetic energies becomes the ratio of the temperatures Tb/Ta = 3/4 since the Boltzmann constant divides out. (KE average = 3/2 kb*T)
c) W = -P*delta_V, but we have to find the pressure first. P = nRT/V = 8315 Pa.
W = -831.5 J
d) using 1st law, Q = delta_U - W = 1.5*(1 mole)(8.315 J/mol*K)(400 K - 300 K) - (-831.5 J) = 2,079 J
2.
a) 0.3 kg/(.004 kg/mol) = 75 moles
b) using 1st law, delta_U = Q + W. We can apply this to the entire process. During the first part, we can assume W = 0 and Q = +3100J. During the second, Q = 0, and W = +1000 J. Thus for the entire process, Q = 3100 J, and W = 1000 J. This means that delta_U = 4100 J.
c) Since delta_U = 4100 J = 1.5nRdelta_T, delta_T = 4.38 K. The final temperature is then 254.4 K.
3. It might help to draw a PV diagram on this problem. There are two segments, and three states to consider.
The initial temperature is 273 K, initial pressure = 101,300 Pa. We can use PV=nRT to find that the initial volume is 2.7 x 10-2 m^3.
At state 2, the volume is 2.7 x 10-2 m^3 (const. volume) and P = 303900 Pa. The temperature is therefore 3*273 K =m 819 K.
At state 3, the volume is 5.4 x 10-2 m^3, and P = 303900 Pa. The temperature at state 3 is therefore 2*819 K = 1638 K.
For this whole process, the work is -(303,900 Pa)(5.4 - 2.7)*10^-3 m^3 = -820.5 J.
The change of internal energy is 3/2*n*R(delta_T) = 17030 J.
From the 1st Law, this means that Q = 17030 J - (-820.5 J) = 17850 J.
Book Problems:
p. 440
32. H = k*A*delta_T/L = (0.84 J/s · m · C°)(3.0 m^2)[15°C - (- 5°C)]/(3.2 x 10-3 m) = 1.6 x 104 W.
p. 471 - 472:
Q3. It is enough information since the process is isothermal - delta_U = 0, so Q = -W.
Q6. In an adiabatic compression, Q = 0, and the work is positive. By the first law, this means delta_U is positive, which means delta_T is positive, an indication that temperature increases.
Problems:
6.
a) Volume doesn't change, so W = 0.
b)Q = -265 kJ, W = 0, so delta_U = -265,000 J.
7.
a) adiabatic means Q = 0.
b) Q = 0, W = 1350 J, so delta_U = 1350 J.
c) since delta_U is positive, the temperature must increase.
8.
a) Work is done only during the constant pressure process, and is equal to -P*delta_V = -1300 J.
b) Q = delta_U - W but since the final temperature is the same as the original, delta_U = 0.
Q = 0 - (-1300 J) = 1300 J.
9.
a) W = -3500 J
b)since the ttemperatures are equal, delta_U = 0.
c) Q = delta_U - W = 0 - (-3500 J) = 3500 J, which represents a heat flow INTO the gas.
1.
a) Tf = 400 K
b) The ratio of average kinetic energies becomes the ratio of the temperatures Tb/Ta = 3/4 since the Boltzmann constant divides out. (KE average = 3/2 kb*T)
c) W = -P*delta_V, but we have to find the pressure first. P = nRT/V = 8315 Pa.
W = -831.5 J
d) using 1st law, Q = delta_U - W = 1.5*(1 mole)(8.315 J/mol*K)(400 K - 300 K) - (-831.5 J) = 2,079 J
2.
a) 0.3 kg/(.004 kg/mol) = 75 moles
b) using 1st law, delta_U = Q + W. We can apply this to the entire process. During the first part, we can assume W = 0 and Q = +3100J. During the second, Q = 0, and W = +1000 J. Thus for the entire process, Q = 3100 J, and W = 1000 J. This means that delta_U = 4100 J.
c) Since delta_U = 4100 J = 1.5nRdelta_T, delta_T = 4.38 K. The final temperature is then 254.4 K.
3. It might help to draw a PV diagram on this problem. There are two segments, and three states to consider.
The initial temperature is 273 K, initial pressure = 101,300 Pa. We can use PV=nRT to find that the initial volume is 2.7 x 10-2 m^3.
At state 2, the volume is 2.7 x 10-2 m^3 (const. volume) and P = 303900 Pa. The temperature is therefore 3*273 K =m 819 K.
At state 3, the volume is 5.4 x 10-2 m^3, and P = 303900 Pa. The temperature at state 3 is therefore 2*819 K = 1638 K.
For this whole process, the work is -(303,900 Pa)(5.4 - 2.7)*10^-3 m^3 = -820.5 J.
The change of internal energy is 3/2*n*R(delta_T) = 17030 J.
From the 1st Law, this means that Q = 17030 J - (-820.5 J) = 17850 J.
Book Problems:
p. 440
32. H = k*A*delta_T/L = (0.84 J/s · m · C°)(3.0 m^2)[15°C - (- 5°C)]/(3.2 x 10-3 m) = 1.6 x 104 W.
p. 471 - 472:
Q3. It is enough information since the process is isothermal - delta_U = 0, so Q = -W.
Q6. In an adiabatic compression, Q = 0, and the work is positive. By the first law, this means delta_U is positive, which means delta_T is positive, an indication that temperature increases.
Problems:
6.
a) Volume doesn't change, so W = 0.
b)Q = -265 kJ, W = 0, so delta_U = -265,000 J.
7.
a) adiabatic means Q = 0.
b) Q = 0, W = 1350 J, so delta_U = 1350 J.
c) since delta_U is positive, the temperature must increase.
8.
a) Work is done only during the constant pressure process, and is equal to -P*delta_V = -1300 J.
b) Q = delta_U - W but since the final temperature is the same as the original, delta_U = 0.
Q = 0 - (-1300 J) = 1300 J.
9.
a) W = -3500 J
b)since the ttemperatures are equal, delta_U = 0.
c) Q = delta_U - W = 0 - (-3500 J) = 3500 J, which represents a heat flow INTO the gas.
Sunday, December 14, 2008
In the Thermodynamic Process of Decorating
Handout solutions:
5. We need to convert to Pascals and m^3: 4.0 L = 4 x 10-3m^3 and 2.0 L = 4 x 10-3m^3.
Since the initial gauge pressure is 3 atm, the absolute pressure is 4 atm = 405,200 Pa.
For a constant pressure process, W = -P*delta_V = -(405,200 Pa)(-2 x 10-3m^3) = 810.4 J
b) Since it's a constant pressure process, the final pressure is also 405,200 Pa.
c) Using ideal gas law, T2 = T1*(V2/V1) = 149 K
6.
a) Using ideal gas law, 5P0*3V0 = nRT0 so T0 = 15P0V0/nR
b) Since temperature is constant, the change of internal energy is zero.
c) Using ideal gas law again, P2V2 = nRT2
We know that T2 = T0 = 15P0V0/nR so P2V2 = 15P0V0.
V2 = 6V0
P2 = 15P0V0/6V0 = 5/2P0
7.
a) Ideal gas law - n = PV/nT = (7*101,300Pa)(10 x 10-3m^3)/(8.315.J/Mol*K * 500 K) = 1.706 moles
b) The container volume is constant, so the work done is zero.
c) T2 = T1*P2/P1 = 214.3 K
d) THe graph should be a vertical line at V = 10 x 10-3 m^3 from P = 7 atm to 3 atm (or the equivalent pressures in Pa.)
Book HW:
3,4,10abc, 53a,c
3.
4.
10
a) and c) PV diagram:
b) W = -2700 J, found since the process is isobaric, using W = -P*delta_V
53.
a) W = -2.2 x 105 J
c)
5. We need to convert to Pascals and m^3: 4.0 L = 4 x 10-3m^3 and 2.0 L = 4 x 10-3m^3.
Since the initial gauge pressure is 3 atm, the absolute pressure is 4 atm = 405,200 Pa.
For a constant pressure process, W = -P*delta_V = -(405,200 Pa)(-2 x 10-3m^3) = 810.4 J
b) Since it's a constant pressure process, the final pressure is also 405,200 Pa.
c) Using ideal gas law, T2 = T1*(V2/V1) = 149 K
6.
a) Using ideal gas law, 5P0*3V0 = nRT0 so T0 = 15P0V0/nR
b) Since temperature is constant, the change of internal energy is zero.
c) Using ideal gas law again, P2V2 = nRT2
We know that T2 = T0 = 15P0V0/nR so P2V2 = 15P0V0.
V2 = 6V0
P2 = 15P0V0/6V0 = 5/2P0
7.
a) Ideal gas law - n = PV/nT = (7*101,300Pa)(10 x 10-3m^3)/(8.315.J/Mol*K * 500 K) = 1.706 moles
b) The container volume is constant, so the work done is zero.
c) T2 = T1*P2/P1 = 214.3 K
d) THe graph should be a vertical line at V = 10 x 10-3 m^3 from P = 7 atm to 3 atm (or the equivalent pressures in Pa.)
Book HW:
3,4,10abc, 53a,c
3.
4.
10
a) and c) PV diagram:
b) W = -2700 J, found since the process is isobaric, using W = -P*delta_V
53.
a) W = -2.2 x 105 J
c)
Thursday, December 11, 2008
In Theory, We Trust - Kinetic Theory Solutions
A bonus assignment for tonight - if you check your answers, please log in and write a comment (go to the bottom of the post for the link) stating the name of your favorite ideal gas. If you don't have one, choose Helium.
50. The gas starts at 0 degrees C, or 273 K. Since v = (3RT/m)^1/2, in order to double v, the temperature must be multiplied by 4. Thus the temperature must be 1092 K.
52. Following the similar logic as with #50, If the pressure doubles, and volume stays the same, the ideal gas equation says that the temperature will also double. This will make the vrms increase by a factor of the square root of 2.
55. You could get to the solution of this equation by looking at the derivation we did today. We ended up with the fact that P = (1/3)m*N*v^3/(V).
The quantity m*N is the total mass of the gas contained in the container, as it's the mass of a single molecule times the number of molecules. The quantity M*N/V must then bethe density of the gas.
Also from the derivation, the v in our expression is the mean square speed of the molecules in the gas.
Thus P = (1/3)*density*v^2
There are other ways to derive it using the other formulas we came up with today, so if your method is different, that is fine.
80. Since you know the temperature, you can directly calculate the vRMS = 183.5 m/s.
Using the result of question 5, we have a relationship between the density, vRMS, and the pressure.
The density will be one atom/cm^3 = 1 amu * 1.66 x 10-27 kg*(106cm^3/m^3) = 1.66 x 10-21 kg/m^3.
Substituting, the pressure is 1.9 x -17 Pa = 1.84 x -22 atm
81. When 10 meters below the surface, the absolute pressure is Patm + rho*g*h = 199,300 Pa.
At the surface, the pressure is 101,300 Pa. You also know the initial volume underneath the water, as well as the fact that the moles of air in the diver's lungs are constant as he/she rises to the surface.
If we assume the rise happens, quickly, we can also say that the temperature is approximately constant.
Thus P1V1/P2V2 = nRT1/nRT2. The entire right side of the equation divides out to 1.
Thus V2 = P1V1/P2 = 10.8 L. I personally would NOT want my lungs stretched to three times their original volume, nor would I expect it to be medically advisable for ANYONE to have this done to them.
This is one of the reasons why it's important for scuba divers to take their time rising to the surface after a dive. One of them is that it causes a condition called 'the bends', where dissolved gas comes out of solution in the blood stream. This is a bad thing. Another is that, without releasing any gas in the lungs during the resurfacing process, the lungs will expand rapidly during the rise, possibly causing tearing in the tissue.
50. The gas starts at 0 degrees C, or 273 K. Since v = (3RT/m)^1/2, in order to double v, the temperature must be multiplied by 4. Thus the temperature must be 1092 K.
52. Following the similar logic as with #50, If the pressure doubles, and volume stays the same, the ideal gas equation says that the temperature will also double. This will make the vrms increase by a factor of the square root of 2.
55. You could get to the solution of this equation by looking at the derivation we did today. We ended up with the fact that P = (1/3)m*N*v^3/(V).
The quantity m*N is the total mass of the gas contained in the container, as it's the mass of a single molecule times the number of molecules. The quantity M*N/V must then bethe density of the gas.
Also from the derivation, the v in our expression is the mean square speed of the molecules in the gas.
Thus P = (1/3)*density*v^2
There are other ways to derive it using the other formulas we came up with today, so if your method is different, that is fine.
80. Since you know the temperature, you can directly calculate the vRMS = 183.5 m/s.
Using the result of question 5, we have a relationship between the density, vRMS, and the pressure.
The density will be one atom/cm^3 = 1 amu * 1.66 x 10-27 kg*(106cm^3/m^3) = 1.66 x 10-21 kg/m^3.
Substituting, the pressure is 1.9 x -17 Pa = 1.84 x -22 atm
81. When 10 meters below the surface, the absolute pressure is Patm + rho*g*h = 199,300 Pa.
At the surface, the pressure is 101,300 Pa. You also know the initial volume underneath the water, as well as the fact that the moles of air in the diver's lungs are constant as he/she rises to the surface.
If we assume the rise happens, quickly, we can also say that the temperature is approximately constant.
Thus P1V1/P2V2 = nRT1/nRT2. The entire right side of the equation divides out to 1.
Thus V2 = P1V1/P2 = 10.8 L. I personally would NOT want my lungs stretched to three times their original volume, nor would I expect it to be medically advisable for ANYONE to have this done to them.
This is one of the reasons why it's important for scuba divers to take their time rising to the surface after a dive. One of them is that it causes a condition called 'the bends', where dissolved gas comes out of solution in the blood stream. This is a bad thing. Another is that, without releasing any gas in the lungs during the resurfacing process, the lungs will expand rapidly during the rise, possibly causing tearing in the tissue.
Wednesday, December 10, 2008
Expand your Mind, and Think of an Ideal Gas - Solutions
p.413, 12, 13, 34, 35, 38, 40 (What is V2/V1?)
12. Notice that you can do this problem without converting to meters, since you end up dividing delta_L by L0, which divides out the units anyway.
You should be able to find that delta_T = delta_L/(alpha*L0) = -89 degrees. (Remember change is final minus initial!) Adding this to the initial temperature of 20 degrees Celsius gives the final temperature of -69 degrees.
13. The key here is multiplying the final lengths of the plate after expansion.
(L + dL)(W + dW) = LW + W*dL + L*dW + dL*dW
As I mentioned in class, the last term is a higher order term, and can be neglected.
If we then substitute using the thermal expansion equation for dW = alpha*W*dT and dL = alpha*L*dT, we get that the final area is LW + 2*alpha*LW*dT.
The question asks for the change in area, so we just have to subtract the initial area LW to get what the text says should be the answer, 2*alpha*L*W*dT.
34. The first task is to determine the number of moles of the gas. Since there are 18.5 kg (18,500 g) of Nitrogen, and the molar weight of Nitrogen gas is 28 g/mol, we can divide to get that there are 661 moles of N2 gas in the tank.
We now have enough information to solve for V = nRT/P (remember that standard temperature and pressure means 273 K, 1.013 x 105 Pa)
V = 14.8 m^3.
For part b, the tank has not changed, so it has the same volume as you found for the first part. We do know there is a new number of moles in the tank (661 + additional 536 moles).
Since we are looking for pressure, we again must solve for P in the ideal gas equation.
P = nRT/V = 1.84 x 105 Pa
35. Remember we have to use absolute pressure, so the initial pressure in the container is (1 atm + .35 atm)*(1.013 x 105 Pa/atm) = 1.37 x 105 Pa
Solving, V = nRT/P = 0.439 m^3
b) Here we are changing states for the gas, so we need to use the division trick from class.
P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give
(P2/P1)(V2/V1) = T2/T1
Solving, T2 = 210 K = -63 degrees Celsius.
38. IN this case, since it is a change of state problem, we do NOT have to convert the pressure to Pascals before doing anything. We can use the same division method as before.
(P2/P1)(V2/V1) = T2/T1
P2 = (T2/T1)(V1/V2)*P1
P2 = (323.2 K/291.2 K)*(55.0 L/48.8 L)* 1 atm = 3.03 atm
40. Again, using the same method:
(P2/P1)(V2/V1) = T2/T1
(V2/V1) = (T2/T1)*(P1/P2) = 1.4
12. Notice that you can do this problem without converting to meters, since you end up dividing delta_L by L0, which divides out the units anyway.
You should be able to find that delta_T = delta_L/(alpha*L0) = -89 degrees. (Remember change is final minus initial!) Adding this to the initial temperature of 20 degrees Celsius gives the final temperature of -69 degrees.
13. The key here is multiplying the final lengths of the plate after expansion.
(L + dL)(W + dW) = LW + W*dL + L*dW + dL*dW
As I mentioned in class, the last term is a higher order term, and can be neglected.
If we then substitute using the thermal expansion equation for dW = alpha*W*dT and dL = alpha*L*dT, we get that the final area is LW + 2*alpha*LW*dT.
The question asks for the change in area, so we just have to subtract the initial area LW to get what the text says should be the answer, 2*alpha*L*W*dT.
34. The first task is to determine the number of moles of the gas. Since there are 18.5 kg (18,500 g) of Nitrogen, and the molar weight of Nitrogen gas is 28 g/mol, we can divide to get that there are 661 moles of N2 gas in the tank.
We now have enough information to solve for V = nRT/P (remember that standard temperature and pressure means 273 K, 1.013 x 105 Pa)
V = 14.8 m^3.
For part b, the tank has not changed, so it has the same volume as you found for the first part. We do know there is a new number of moles in the tank (661 + additional 536 moles).
Since we are looking for pressure, we again must solve for P in the ideal gas equation.
P = nRT/V = 1.84 x 105 Pa
35. Remember we have to use absolute pressure, so the initial pressure in the container is (1 atm + .35 atm)*(1.013 x 105 Pa/atm) = 1.37 x 105 Pa
Solving, V = nRT/P = 0.439 m^3
b) Here we are changing states for the gas, so we need to use the division trick from class.
P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give
(P2/P1)(V2/V1) = T2/T1
Solving, T2 = 210 K = -63 degrees Celsius.
38. IN this case, since it is a change of state problem, we do NOT have to convert the pressure to Pascals before doing anything. We can use the same division method as before.
(P2/P1)(V2/V1) = T2/T1
P2 = (T2/T1)(V1/V2)*P1
P2 = (323.2 K/291.2 K)*(55.0 L/48.8 L)* 1 atm = 3.03 atm
40. Again, using the same method:
(P2/P1)(V2/V1) = T2/T1
(V2/V1) = (T2/T1)*(P1/P2) = 1.4
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