Question 18.8: In a fiendish plan to incapacitate Superman, the arch villai...
In a fiendish plan to incapacitate Superman, the arch villain Dorkmann proposes to compress 3.50 kg of krypton gas from 1.00 atmosphere, 20.0°C to 10.0 MPa, producing a concentrated and possibly toxic concentration of kryptonite. The foolish, unschooled fiend intends to try to carry out the compression process adiabatically using only 100. kJ of work. But you, as the ever-present hero Thermoperson, seeker of truth and wisdom, hold the power to foil the plan by computing
a. The final temperature of the krypton gas after compression.
b. The entropy production of the compression process.
a. The final temperature of the gas can be found from an energy balance and Eq. (18.46a) as
u = \frac{3}{2} RT (18.46a)
_1Q_2 − _1W_2 = m(u_2 − u_1) = m (\frac{3}{2} R) (T_2 − T_1)
where R_\text{krypton} = ℜ/M_\text{krypton} = 8.3143/83.80 = 0.0992 \text{kJ/kg.K}. Since _1Q_2 = 0 for an adiabatic compression, the energy balance equation can then be solved for T_2 as
T_2 = T_1 − \frac{_1W_2}{3mR/2} = (20.0 + 273.15 \text{K}) − \frac{−100. \text{kJ}}{3(3.50 \text{kg})(0.0992 \text{kJ/kg.K})/2} = 458 \text{K}
b. An entropy balance using Eq. (18.46e) produces
s = R \left\{\text{ln} [(2πm/ħ^2)^{3/2} (kT)^{5/2} /p] + \frac{5}{2}\right\} (18.46e)
\frac{_1Q_2}{T_b} + _1(S_P)_2 = m(s_2 − s_1)= m(R \left\{\text{ln} [\frac{(2πm/ħ^2)^{3/2} (kT_2)^{5/2}}{p_2} + \frac{5}{2}]\right\} – R \left\{\text{ln} [\frac{(2πm/ħ^2)^{3/2} (kT_1)^{5/2}}{p_1}] + \frac{5}{2}\right\} )
Solving this equation for _1(S_P)_2 gives
_1(S_P)_2 = mR \text{ln} [(\frac{T_2}{T_1})^{5/2} (\frac{p_1}{p_2})]
= (3.50 \text{kg}) (0.0992 \text{kJ/kg.K}) \text{ln} [(\frac{485 \text{K}}{293.15 \text{K}})^{5/2} (\frac{0.101325 \text{MPa}}{10.0 \text{MPa}})] = −1.16 \frac{\text{kJ}}{\text{kg.K}}
Since _1(S_P)_2 is less than zero here, it violates the second law of thermodynamics and this process can not possibly occur. Therefore, Superman has nothing to worry about.