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.