MikaStars39/nano-eval huggingface dataset:
The relationship between the standard equilibrium constant and temperature for the reaction $\mathrm{CO} (\mathrm{g}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} (\mathrm{g}) + \mathrm{CO}_{2} (\mathrm{g})$ is given by the equation: $\lg K_{p}^{\Theta} = \frac{2150 \, \mathrm{K}}{T} - 2.216$. Given the initial mass fractions of $\mathrm{CO}$, $\mathrm{H}_{2}\mathrm{O}$, $\mathrm{H}_{2}$, and $\mathrm{CO}_{2}$ are 0.30, 0.30, 0.20, and 0.20 respectively, and the total pressure is 100.00 kPa, at what temperature will the reaction proceed towards the formation of products?

Options:
A. $$
T < 8 3 0 \, \mathrm{~ K}
$$
B. $$
T < 5 0 0 \, \mathrm{~ K}
$$
C. $$
T < 7 0 0 \, \mathrm{~ K}
$$
D. $$
T < 9 0 0 \, \mathrm{~ K}
$$
E. $$
T > 1 0 0 0 \, \mathrm{~ K}
$$
F. $$
T = 8 5 0 \, \mathrm{~ K}
$$
G. $$
T = 1 0 0 0 \, \mathrm{~ K}
$$
H. $$
T > 1 5 0 0 \, \mathrm{~ K}
$$
I. $$
T = 5 0 0 \, \mathrm{~ K}
$$
J. $$
T > 8 3 0 \, \mathrm{~ K}
$$
Answer: A
The reaction between $\mathrm{CO}_{2} \left( \mathrm{g} \right)$ and $\mathrm{H}_{2} \mathrm{S} \left( \mathrm{g} \right)$ at high temperature is $\mathrm{CO}_{2} \left( \mathrm{g} \right) + \mathrm{H}_{2} \mathrm{S} \left( \mathrm{g} \right) = \mathrm{COS} \left( \mathrm{g} \right) + \mathrm{H}_{2} \mathrm{O} \left( \mathrm{g} \right)$. Now, when 4.4 g of $\mathrm{CO}_{2} \left( \mathrm{g} \right)$ is added to an empty container with a volume of $2.5 \, \mathrm{dm}^{3}$ at 610 K, and then $\mathrm{H}_{2} \mathrm{S} \left( \mathrm{g} \right)$ is introduced to reach a total pressure of 1000 kPa, the equilibrium reached and the sample analysis shows that the mole fraction of $\mathrm{H}_{2} \mathrm{O} \left( \mathrm{g} \right)$ is 0.02. Repeating the experiment by raising the temperature to 620 K and reaching equilibrium again shows the mole fraction of $\mathrm{H}_{2} \mathrm{O} \left( \mathrm{g} \right)$ to be 0.03. Considering the gas as an ideal gas, calculate the $K_{p}^{\Theta}$ at 610 K.

Options:
A. $$
3. 5 8 \times1 0^{-3}
$$
B. $$
5. 4 1 \times1 0^{-4}
$$
C. $$
7. 1 0 \times1 0^{-4}
$$
D. $$
2. 5 6 \times1 0^{-2}
$$
E. $$
9. 0 1 \times1 0^{-4}
$$
F. $$
1. 2 3 \times1 0^{-2}
$$
G. $$
8. 9 7 \times1 0^{-3}
$$
H. $$
4. 3 3 \times1 0^{-5}
$$
I. $$
1. 8 2 \times1 0^{-2}
$$
J. $$
6. 6 9 \times1 0^{-3}
$$
Answer: J
For the following ideal gas reaction $\mathrm{A}_{2} \left( \mathrm{g ~} \right) \,+\, \mathrm{B}_{2} \left( \mathrm{ g ~} \right)=2 \, \mathrm{A B} \left( \mathrm{ g ~} \right)$, if the molar masses of A and B are approximately equal, $M_{\mathrm{A}} \approx M_{\mathrm{B}} \approx 36 \mathrm{\ g \ \cdot\ m o l^{-1}}$, and the atomic distances in the molecules $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and AB are nearly the same $r_{\mathrm{A A}} \approx r_{\mathrm{B B}} \approx r_{\mathrm{A B}} \approx 10^{-10} \mathrm{~ m}$, with their vibrational frequencies also nearly equal, $\nu_{\mathrm{A A}} \, \approx\, \nu_{\mathrm{B B}} \, \approx\, \nu_{\mathrm{A B}} \, \approx\, 10^{13} \, \mathrm{s}^{-1}$, and the difference in ground state energy between the product molecules and reactant molecules is $\Delta_{\mathrm{r}} U_{\mathrm{m}}^{\Theta} ( 0 ) \; \approx\;-\, 8. \; 3 1 4 \mathrm{~ kJ ~ \cdot~ mol ~}$. Assuming the standard enthalpy change of the reaction is temperature independent and that $\Delta_{\mathrm{r}} H_{\mathrm{m}}^{\Theta} \approx L \Delta\varepsilon_{0} \approx\ \Delta_{\mathrm{r}} U_{\mathrm{m}}^{\Theta} \ ( 0 )$, find the change in reaction entropy $\Delta_{\mathrm{r}} S_{\mathrm{m}}^{\Theta}$.

Options:
A. $$
1 2. 9 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
B. $$
9. 8 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
C. $$
8. 0 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
D. $$
6. 7 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
E. $$
5. 6 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
F. $$
2 3. 1 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
G. $$
4. 2 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
H. $$
7. 4 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
I. $$
3. 9 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
J. $$
1 1. 5 \mathrm{~ J \cdot~ m o l^{-1} ~ \cdot~ K^{-1} ~}
$$
Answer: J
Based on the following data, determine the degree of dissociation and the ionic product constant $K_{\mathrm{w}}^{\Theta}$. For $\mathrm{H}_{2} \, \mathrm{O} (l)$ when it dissociates into H⁺ and OH⁻ and reaches equilibrium at 298 K. It is known that at 298 K, the conductivity of pure water is $\kappa_{\mathrm{H_{2}O}} = 5.5 \times 10^{-6} \, \mathrm{S \cdot m^{-1}}$, the molar conductivity of $H^+$ at infinite dilution $A_{\mathrm{m, H^+}}^{\mathrm{\infty}} = 3.498 \times 10^{-2} \, \mathrm{S \cdot m^2 \cdot mol^{-1}}$, and the molar conductivity of $OH^-$ at infinite dilution $\varLambda_{\mathrm{m, OH^-}}^{\mathrm{\infty}} = 1.98 \times 10^{-2} \, \mathrm{S \cdot m^2 \cdot mol^{-1}}$. The density of water is 997.09 $\mathrm{kg \cdot m^{-3}}$.

Options:
A. $$
6. \, 2 3 1 \, \times\, 1 0^{-3} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
B. $$
7. \, 1 9 4 \, \times\, 1 0^{-3} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
C. $$
3. \, 8 4 7 \, \times\, 1 0^{-5} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
D. $$
8. \, 6 9 2 \, \times\, 1 0^{-3} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
E. $$
5. \, 4 7 8 \, \times\, 1 0^{-2} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
F. $$
9. \, 1 5 2 \, \times\, 1 0^{-4} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
G. $$
2. \, 0 8 1 \, \times\, 1 0^{-4} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
H. $$
4. \, 5 7 3 \, \times\, 1 0^{-5} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
I. $$
1. \, 3 6 9 \, \times\, 1 0^{-6} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
J. $$
4. \, 9 5 2 \, \times\, 1 0^{-4} \, \, \mathrm{S} \, \cdot\, \mathrm{m}^{2} \, \cdot\, \mathrm{m o l}^{-1}
$$
Answer: E
The molal concentration of a dilute solution is $m$, the boiling point elevation is $\Delta T_{\mathrm{b}}$, and the freezing point depression is $\Delta T_{\mathrm{f}}$, then_____.

Options:
A. $$
\Delta\, T_{\mathrm{b}} < 0
$$
B. $$
\Delta\, T_{\mathrm{f}} < \Delta\, T_{\mathrm{b}}
$$
C. $$
\Delta\, T_{\mathrm{b}} = m
$$
D. $$
\Delta\, T_{\mathrm{b}} = \Delta\, T_{\mathrm{f}}
$$
E. $$
\Delta\, T_{\mathrm{b}} > 0
$$
F. $$
\Delta\, T_{\mathrm{f}} = m
$$
G. $$
\Delta\, T_{\mathrm{f}} > \Delta\, T_{\mathrm{b}}
$$
H. $$
\Delta\, T_{\mathrm{f}} = 0
$$
I. $$
\Delta\, T_{\mathrm{f}} = 1
$$
J. $$
\Delta\, T_{\mathrm{f}} > 0
$$
Answer: G
Initial state of the gas: $p_{1}=0.5 \mathrm{~MPa}$, $V_{1}=0.4 \mathrm{~m}^{3}$, and it expands to $V_{2}=0.8 \mathrm{~m}^{3}$ under constant pressure conditions. Calculate the expansion work done by the system during the process

Options:
A. $$
8 5. 2 5 k J
$$
B. $$
4 2. 6 2 k J
$$
C. $$
7 5. 9 5 k J
$$
D. $$
6 4. 8 2 k J
$$
E. $$
1 0 2 . 3 k J
$$
F. $$
3 8. 5 0 k J
$$
G. $$
7 0. 5 0 k J
$$
H. $$
1 2. 4 8 k J
$$
I. $$
5 3. 7 5 k J
$$
J. $$
9 0. 3 1 k J
$$
Answer: A
An insulated container with a partition separates two sections, one containing $n_{1}$ moles of an ideal gas at pressure $p_{1}$, and the other containing $n_{2}$ moles of an ideal gas at pressure $p_{2}$. Both sides are at the same temperature $T$. If the partition is removed and the gases on both sides are different types, what is the change in entropy after mixing?

Options:
A. $$
p_{2}^{\prime}=\frac{p_{1} V_{2}} {V_{2}+V_{1}};~\Delta S=n_{1} R \operatorname{l n} {\frac{V_{2}+V_{1}} {V_{2}}}+n_{2} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}
$$
B. $$
p_{1}^{\prime}=\frac{p_{2} V_{2}} {V_{1}+V_{2}};~\Delta S=n_{2} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}+n_{1} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{2}}}
$$
C. $$
p_{1}^{\prime}=\frac{p_{1} V_{2}} {V_{1}-V_{2}};~\Delta S=n_{2} R \operatorname{l n} {\frac{V_{1}-V_{2}} {V_{2}}}+n_{1} R \operatorname{l n} {\frac{V_{1}-V_{2}} {V_{1}}}
$$
D. $$
p_{1}^{\prime}=\frac{p_{2} V_{1}} {V_{1}+V_{2}};~\Delta S=n_{2} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}+n_{1} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{2}}}
$$
E. $$
p_{2}^{\prime}=\frac{p_{2} V_{2}} {V_{1}+V_{2}};~\Delta S=n_{1} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}+n_{2} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{2}}}
$$
F. $$
p_{1}^{\prime}=\frac{p_{1} V_{1}} {V_{2}-V_{1}};~\Delta S=n_{2} R \operatorname{l n} {\frac{V_{2}-V_{1}} {V_{1}}}+n_{1} R \operatorname{l n} {\frac{V_{2}-V_{1}} {V_{2}}}
$$
G. $$
p_{2}^{\prime}=\frac{p_{1} V_{2}} {V_{1}+V_{2}};~\Delta S=n_{2} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{2}}}+n_{1} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}
$$
H. $$
p_{2}^{\prime}=\frac{p_{2} V_{1}} {V_{2}+V_{1}};~\Delta S=n_{1} R \operatorname{l n} {\frac{V_{2}+V_{1}} {V_{2}}}+n_{2} R \operatorname{l n} {\frac{V_{2}+V_{1}} {V_{1}}}
$$
I. $$
p_{1}^{\prime}=\frac{p_{2} V_{2}} {V_{2}+V_{1}};~\Delta S=n_{2} R \operatorname{l n} {\frac{V_{2}+V_{1}} {V_{2}}}+n_{1} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}
$$
J. $$
p_{2}^{\prime}=\frac{p_{1} V_{1}} {V_{1}+V_{2}};~\Delta S=n_{1} R \operatorname{l n} {\frac{V_{1}+V_{2}} {V_{1}}}+n_{2} R \operatorname{l n} {\frac{V_{1}-V_{2}} {V_{2}}}
$$
Answer: E
The atmosphere is often in a convective stable state of equal entropy. In this equilibrium, $p \upsilon^{\gamma}$ does not change with height, where $\gamma = c_{p} / c$. Using the condition of hydrostatic equilibrium in a uniform gravitational field, find $\frac{d T} {d z}$ as ___.

Options:
A. $$
- \frac{\gamma-1} {\gamma} \cdot \frac{g m} {k}
$$
B. $$
- \frac{\gamma-1} {\gamma} \cdot\frac{g k} {m}
$$
C. $$
- \frac{\gamma-1} {\gamma} \cdot\frac{m g} {k}
$$
D. $$
- \frac{\gamma} {\gamma-1} \cdot\frac{m g} {k}
$$
E. $$
- \frac{\gamma-1} {\gamma} \cdot \frac{k} {m g}
$$
F. $$
- \frac{\gamma} {\gamma-1} \cdot\frac{k} {m g}
$$
G. $$
- \frac{\gamma+1} {\gamma} \cdot \frac{m g} {k}
$$
H. $$
- \frac{\gamma-1} {\gamma} \cdot\frac{m} {g k}
$$
I. $$
- \frac{\gamma-1} {\gamma+1} \cdot \frac{m g} {k}
$$
J. $$
- \frac{1-\gamma} {\gamma} \cdot \frac{m g} {k}
$$
Answer: C
Fermi Gas. Consider an ideal Fermi gas, with atomic mass $m=5 \times 10^{-26} \text{g}$, nuclear spin $I=\frac{1}{2}$, and nuclear magnetic moment $\mu=1 \times 10^{-25} \text{erg/G s}$. When $T=0 \text{K}$, what is the maximum density at which the gas can be fully polarized by an external magnetic field of $10^{5} \, \text{Gs}$? (Assuming no electronic magnetic moment.)

Options:
A. $\approx2 \times 10^{17}$ atoms/$cm^{3}$
B. $\approx7 \times 10^{15}$ atoms/$cm^{3}$
C. $\approx3 \times 10^{18}$ atoms/$cm^{3}$
D. $\approx2 \times 10^{14}$ atoms/$cm^{3}$
E. $\approx6 \times 10^{16}$ atoms/$cm^{3}$
F. $\approx5 \times 10^{16}$ atoms/$cm^{3}$
G. $\approx4 \times 10^{17}$ atoms/$cm^{3}$
H. $\approx9 \times 10^{18}$ atoms/$cm^{3}$
I. $\approx8 \times 10^{15}$ atoms/$cm^{3}$
J. $\approx1 \times 10^{19}$ atoms/$cm^{3}$
Answer: A
In an air conditioning system, air flows through a horizontal pipeline of constant diameter, with no heat exchange with the external environment. At cross section I, the pressure is 150 psia, the flow speed is 80 ft/sec, and the temperature is 70°F. At section 2, the pressure is 120 psia, and the temperature is 50°F. All changes refer to the variation between sections 1 and 2. Assuming the air is an ideal gas, what is the change of enthalpy ?

Options:
A. $$
1 5 4 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
B. $$
1 4 6 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
C. $$
1 5 5 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
D. $$
1 4 8 5 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
E. $$
1 5 0 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
F. $$
1 4 7 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
G. $$
1 5 1 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
H. $$
1 4 9 3 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
I. $$
1 5 3 0 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
J. $$
1 5 2 5 \mathrm{f t} \cdot\mathrm{l b f} / \mathrm{s l u g}
$$
Answer: H
The oil with a viscosity of $0.0008 \, \mathrm{lb \cdot sec/ft^2}$ (specific gravity $s=0.90$) flows at a rate of $2.0 \, \mathrm{ft^3/sec}$ through a pipe with a diameter of 6 inches and a length of 200 feet. Its friction head loss is $25 \, \mathrm{ft \cdot lbf/lbf_{\circ}}$. What is the flow velocity at the centerline of the pipe?

Options:
A. $$
1 4. 6 2 {\mathrm{ft}} / \mathrm{s e c}
$$
B. $$
1 2. 8 5 {\mathrm{ft}} / \mathrm{s e c}
$$
C. $$
2 0. 3 0 {\mathrm{ft}} / \mathrm{s e c}
$$
D. $$
1 5. 0 0 {\mathrm{ft}} / \mathrm{s e c}
$$
E. $$
1 1. 2 5 {\mathrm{ft}} / \mathrm{s e c}
$$
F. $$
2 2. 0 5 {\mathrm{ft}} / \mathrm{s e c}
$$
G. $$
1 8. 9 7 {\mathrm{ft}} / \mathrm{s e c}
$$
H. $$
0 7. 4 3 {\mathrm{ft}} / \mathrm{s e c}
$$
I. $$
0 9. 8 5 {\mathrm{ft}} / \mathrm{s e c}
$$
J. $$
1 0. 7 5 {\mathrm{ft}} / \mathrm{s e c}
$$
Answer: B
In a circular rectangular channel, the water depth is $0.5 m$ and the flow velocity is $6 \mathrm{m/s}$. What is the percentage by which the specific energy exceeds the minimum value?

Options:
A. $$
6 0. 2 \%
$$
B. $$
4 8. 3 \%
$$
C. $$
5 1. 0 \%
$$
D. $$
7 6. 9 \%
$$
E. $$
7 2. 5 \%
$$
F. $$
6 3. 1 \%
$$
G. $$
5 9. 4 \%
$$
H. $$
5 5. 7 \%
$$
I. $$
6 7. 8 \%
$$
J. $$
8 1. 3 \%
$$
Answer: A
In a pipeline with a diameter of 3in, a flow nozzle with a diameter of 2in is installed, allowing water to flow through at $72 ^\circ F$. If a water-air pressure gauge shows a difference of 2in, what is the volume of the flow?

Options:
A. $$
0. 1 0 5 9\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
B. $$
0. 0 6 9 4\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
C. $$
0. 0 7 3 2\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
D. $$
0. 0 4 6 8\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
E. $$
0. 1 4 3 0\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
F. $$
0. 1 1 5 2\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
G. $$
0. 1 2 7 3\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
H. $$
0. 0 2 8 5\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
I. $$
0. 0 8 8 0\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
J. $$
0. 0 9 5 6\ \mathrm{f t}^{3} / \mathrm{s e c}
$$
Answer: C
Under the condition of 80°F, gas in a large container flows out through a nozzle with an inside diameter of 20 inches, exhausting into an atmosphere with a pressure of 13.5 psia. What are the mass flow rate through the nozzle when the container pressure is 5 psig, 15 psig, and 20 psig. Assume that under isentropic conditions, $m$ is a function of $p_{0}$, and that the temperature of the container is 80°F under any condition.

Options:
A. $$
0. 1 0 4 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 3 2 8 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 9 4 5 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
B. $$
0. 0 5 1 3 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 4 6 8 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 8 0 2 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
C. $$
0. 0 4 8 9 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 4 2 5 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 6 9 3 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
D. $$
0. 0 2 9 5 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 5 8 3 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 8 4 5 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
E. $$
0. 0 6 7 2 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 3 1 7 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 4 5 8 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
F. $$
0. 0 3 2 4 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 5 6 2 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 8 1 7 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
G. $$
0. 0 4 1 8 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 2 4 9 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 6 1 2 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
H. $$
0. 0 374 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 6 3 4 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 745 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
I. $$
0. 0 3 5 2 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 7 8 4 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 5 2 9 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
J. $$
0. 0 5 3 9 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 4 7 6 {\mathrm{s l u g}} / {\mathrm{s e c}},\
0. 0 5 7 8 {\mathrm{s l u g}} / {\mathrm{s e c}}
$$
Answer: H
The diameter of the impulse turbine is 10.0 ft (3.05 m), with a bucket angle $\beta_{2}=160^{\circ}$. If the nozzle jet velocity $V_{1}$ = 200 ft/sec (61 m/s), nozzle diameter = 2.0 in (5.08 cm), and rotational speed $= 240 \mathrm{r/min}$, what is the power of the impeller? Assume that the velocity of the liquid relative to the bucket remains constant (i.e., $v_{1}=v_{2}$).

Options:
A. $$
1 9 5. 3 h p
$$
B. $$
3 0 1. 0 h p
$$
C. $$
2 3 7. 5 h p
$$
D. $$
3 0 9. 8 h p
$$
E. $$
2 7 8. 6 h p
$$
F. $$
2 9 0. 7 h p
$$
G. $$
3 1 5. 1 h p
$$
H. $$
2 6 7. 4 h p
$$
I. $$
2 4 9. 9 h p
$$
J. $$
2 5 8. 2 h p
$$
Answer: E
A silicon sample with a conductivity of $0.001 / ( \Omega \cdot \mathrm{cm} )$ has a Hall voltage of zero under a weak magnetic field. Assuming that the Hall factors for electrons [ $\mu_{\mathrm{n}}=1300 \mathrm{cm}^{2} / (\mathrm{V} \cdot \mathrm{s})$ ] and holes [ $\mu_{p}=300 \mathrm{cm}^{2} / (\mathrm{V} \cdot \mathrm{s})$ ] are the same, what is the carrier density?

Options:
A. $$
6. 1 1 \times1 0^{1 1} ( 1 / \mathrm{c m}^{3} )
$$
B. $$
2. 7 3 \times1 0^{1 1} ( 1 / \mathrm{c m}^{3} )
$$
C. $$
3. 0 5 \times1 0^{1 3} ( 1 / \mathrm{c m}^{3} )
$$
D. $$
4. 3 0 \times1 0^{1 4} ( 1 / \mathrm{c m}^{3} )
$$
E. $$
1. 6 9 \times1 0^{1 3} ( 1 / \mathrm{c m}^{3} )
$$
F. $$
7. 9 0 \times1 0^{1 3} ( 1 / \mathrm{c m}^{3} )
$$
G. $$
2. 4 5 \times1 0^{1 2} ( 1 / \mathrm{c m}^{3} )
$$
H. $$
5. 6 7 \times1 0^{1 2} ( 1 / \mathrm{c m}^{3} )
$$
I. $$
8. 1 2 \times1 0^{1 2} ( 1 / \mathrm{c m}^{3} )
$$
J. $$
5. 0 0 \times1 0^{1 4} ( 1 / \mathrm{c m}^{3} )
$$
Answer: E
The thermal conductivity of $\mathrm{Bi}_{2} \mathrm{Te}_{3}$ is $2.4 \, \left[ \mathrm{W/(m \cdot K)} \right]$, while the electrical conductivity of n-type $\mathrm{Bi}_{2} \mathrm{Te}_{3}$ is $10^{5} \, \mathrm{s/m}$. What percentage of the thermal conductivity contribution is made by carriers at $300 \, \mathrm{K}$? (Assume long wavelength acoustic wave scattering)

Options:
A. $$
2 0. 3\%
$$
B. $$
1 6. 9\%
$$
C. $$
2 2. 5\%
$$
D. $$
2 1. 0\%
$$
E. $$
1 9. 5\%
$$
F. $$
1 4. 2\%
$$
G. $$
1 7. 4\%
$$
H. $$
1 5. 8\%
$$
I. $$
1 8. 7\%
$$
J. $$
2 3. 7\%
$$
Answer: I
Consider an $\mathbf{n}$-type semiconductor material with a dark conductivity of 100 S/cm at room temperature. When exposed to light with intensity $I = 10^{-6} \, \mathrm{W / cm}^{2}$, and an absorption coefficient $\alpha = 10^{2} / \mathrm{cm}$, the ratio of the steady-state photoconductivity to the dark conductivity $\gamma = 10$. The carrier lifetime $\tau$ is $10^{-4} \, \mathrm{s}$, and the mobility ratio $b = \mu_{n} / \mu_{p} = 10$, with electron mobility $\mu_{n} = 10,000 \, \mathrm{cm}^{2} / (\mathrm{V \cdot s})$. What is the corresponding quantum efficiency?

Options:
A. $$
8. 9 5 \times1 0^{2 2}
$$
B. $$
2. 4 3 \times1 0^{1 8}
$$
C. $$
6. 3 7 \times1 0^{1 4}
$$
D. $$
3. 6 8 \times1 0^{1 7}
$$
E. $$
5. 1 1 \times1 0^{2 1}
$$
F. $$
7. 8 4 \times1 0^{1 6}
$$
G. $$
0. 2 8 \times1 0^{2 3}
$$
H. $$
9. 1 2 \times1 0^{1 5}
$$
I. $$
1. 7 3 \times1 0^{1 9}
$$
J. $$
4. 5 6 \times1 0^{2 0}
$$
Answer: E
What is the change in flat-band voltage when positive charges are uniformly distributed in the oxide layer?

Options:
A. $$
3. 0 ( \mathrm{V} )
$$
B. $$
4. 4 ( \mathrm{V} )
$$
C. $$
2. 7 ( \mathrm{V} )
$$
D. $$
- 4. 6 ( \mathrm{V} )
$$
E. $$
- 3. 3 ( \mathrm{V} )
$$
F. $$
- 5. 2 ( \mathrm{V} )
$$
G. $$
- 2. 1 ( \mathrm{V} )
$$
H. $$
- 1. 9 ( \mathrm{V} )
$$
I. $$
0. 83 ( \mathrm{V} )
$$
J. $$
1. 5 ( \mathrm{V} )
$$
Answer: D
An ideal MOS capacitor has a substrate of p-type silicon with a doping concentration of $N_{\mathrm{A}} = 1.5 \times 10^{15} / \mathrm{cm}^3$. If the oxide layer thickness is $0.1 \; \mu\mathrm{m}$, the threshold voltage $V_{\mathrm{T}}$ is 1.1 V. What is the value of $V_{\mathrm{T}}$ when the oxide layer thickness is $0.2 \; \mu\mathrm{m}$?

Options:
A. $$
1. 1 ( \mathrm{V} )
$$
B. $$
1. 4 ( \mathrm{V} )
$$
C. $$
1. 9 ( \mathrm{V} )
$$
D. $$
1. 0 ( \mathrm{V} )
$$
E. $$
1. 6 ( \mathrm{V} )
$$
F. $$
1. 2 ( \mathrm{V} )
$$
G. $$
1. 8 ( \mathrm{V} )
$$
H. $$
3. 0 ( \mathrm{V} )
$$
I. $$
2. 7 ( \mathrm{V} )
$$
J. $$
1. 3 ( \mathrm{V} )
$$
Answer: E
A pn junction diode has the following parameters: $N_{\mathrm{D}} = 10^{16} \, \text{cm}^{-3}$, $N_{\mathrm{A}} = 5 \times 10^{18} \, \text{cm}^{-3}$, $\tau_{\mathrm{n}} = \tau_{p} = 1 \, \mu\text{s}$, and $A = 0.01 \, \text{cm}^2$. Assume that the widths on both sides of the junction are much larger than the diffusion lengths of their respective minority carriers. Assume the electron mobility in the p-type region $\mu_{\mathrm{n}} = 500 \, \text{cm}^2 / (\text{V} \cdot \text{s})$, and the hole mobility in the n-type region $\mu_{p} = 180 \, \text{cm}^2 / (\text{V} \cdot \text{s})$. What is the applied voltage at room temperature (300 K) when the forward current is 1 mA?

Options:
A. $$
6 2 0 ( \mathrm{m V} )
$$
B. $$
5 8 5 ( \mathrm{m V} )
$$
C. $$
5 9 0 ( \mathrm{m V} )
$$
D. $$
7 0 0 ( \mathrm{m V} )
$$
E. $$
6 4 5 ( \mathrm{m V} )
$$
F. $$
6 5 5 ( \mathrm{m V} )
$$
G. $$
6 1 0 ( \mathrm{m V} )
$$
H. $$
5 6 0 ( \mathrm{m V} )
$$
I. $$
6 3 5 ( \mathrm{m V} )
$$
J. $$
6 8 0 ( \mathrm{m V} )
$$
Answer: G
An $\mathbf{n}$-type germanium sample has a resistivity of $\rho = 10 \, \Omega \cdot \mathrm{cm}$. Assume the recombination center energy level $E_{\mathrm{t}}$ is 0.22 eV above the valence band top, with $\tau_{\mathrm{n}} = 100 \, \mu\mathrm{s}$ and $\tau_{\mathrm{p}} = 10 \, \mu\mathrm{s}$. What is the lifetime $\tau$ of the sample at room temperature, assuming $N_{\mathrm{c}} \approx N_{\mathrm{r}} = 10^{19} \, \mathrm{cm}^{-3}$?

Options:
A. $$
1 3 1 3 ( \mu\mathrm{s} )
$$
B. $$
2 5 0 0 ( \mu\mathrm{s} )
$$
C. $$
1 9 2 0 ( \mu\mathrm{s} )
$$
D. $$
6 4 9 0 ( \mu\mathrm{s} )
$$
E. $$
9 1 2 2 ( \mu\mathrm{s} )
$$
F. $$
2 1 1 1 ( \mu\mathrm{s} )
$$
G. $$
3 6 4 0 ( \mu\mathrm{s} )
$$
H. $$
2 3 4 5 ( \mu\mathrm{s} )
$$
I. $$
8 7 5 2 ( \mu\mathrm{s} )
$$
J. $$
4 0 0 5 ( \mu\mathrm{s} )
$$
Answer: A
A thermistor made from intrinsic silicon material has a resistance value of 500 $\Omega$ at 290K. Assuming the band gap of silicon $E_{\mathrm{_{o}}}=1.12 \mathrm{eV}$ and considering it does not change with temperature, what is the approximate value of the thermistor at 325K if the carrier mobility is assumed to remain constant?

Options:
A. $$
4 2. 7 ( \Omega)
$$
B. $$
3 9. 0 ( \Omega)
$$
C. $$
3 7. 8 ( \Omega)
$$
D. $$
3 9. 2 ( \Omega)
$$
E. $$
3 6. 5 ( \Omega)
$$
F. $$
3 8. 1 ( \Omega)
$$
G. $$
3 7. 2 ( \Omega)
$$
H. $$
4 0. 5 ( \Omega)
$$
I. $$
3 6. 9 ( \Omega)
$$
J. $$
4 1. 3 ( \Omega)
$$
Answer: F
A silicon material with impurity compensation is given, known to have an acceptor density $N_{\mathrm{A}} = 1 \times 10^{15} / \mathrm{cm}^{3}$. At room temperature, its $E_{\mathrm{F}}$ is measured to coincide precisely with the donor energy level, and the equilibrium electron density is known to be $n_{0} = 5 \times 10^{15} / \mathrm{cm}^{3}$. The intrinsic carrier density of silicon at room temperature is known to be $n_{i} = 1.5 \times 10^{10} / \mathrm{cm}^{3}$. What is the donor impurity density in the doped material?

Options:
A. $$
3. 6 {\times} 1 0^{1 4} \mathrm{c m}^{-3}
$$
B. $$
2. 7 {\times} 1 0^{1 5} \mathrm{c m}^{-3}
$$
C. $$
4. 4 {\times} 1 0^{1 4} \mathrm{c m}^{-3}
$$
D. $$
6. 3 {\times} 1 0^{1 4} \mathrm{c m}^{-3}
$$
E. $$
1. 0 {\times} 1 0^{1 5} \mathrm{c m}^{-3}
$$
F. $$
1. 8 {\times} 1 0^{1 6} \mathrm{c m}^{-3}
$$
G. $$
1. 1 {\times} 1 0^{1 6} \mathrm{c m}^{-3}
$$
H. $$
5. 0 {\times} 1 0^{1 5} \mathrm{c m}^{-3}
$$
I. $$
7. 9 {\times} 1 0^{1 5} \mathrm{c m}^{-3}
$$
J. $$
2. 2 {\times} 1 0^{1 6} \mathrm{c m}^{-3}
$$
Answer: F
If the Fermi level $E_{\mathrm{F}}=5 \mathrm{eV}$, use the Fermi function to calculate the temperature at which the probability of electrons occupying the energy level $E=5.5 \mathrm{eV}$ is 1%. At this temperature, what is the energy range corresponding to the electron distribution probability from 0.9 to 0.1?

Options:
A. $$
0. 5 4 ( \mathrm{e V} )
$$
B. $$
0. 3 9 ( \mathrm{e V} )
$$
C. $$
0. 2 8 ( \mathrm{e V} )
$$
D. $$
1. 5 2 ( \mathrm{e V} )
$$
E. $$
0. 7 1 ( \mathrm{e V} )
$$
F. $$
1. 1 0 ( \mathrm{e V} )
$$
G. $$
0. 6 7 ( \mathrm{e V} )
$$
H. $$
0. 9 5 ( \mathrm{e V} )
$$
I. $$
0. 3 2 ( \mathrm{e V} )
$$
J. $$
0. 4 8 ( \mathrm{e V} )
$$
Answer: J
Steam at $t_s = 40$°C and R134a vapor at $t_{\varepsilon} = 40$°C. Film condensation on an isothermal vertical wall. Try to calculate the liquid film thickness at 0.5m away from $x = 0$. Assume $\Delta t = t_w - t_x = 5$°C.

Options:
A. $$
2. 0 4 6 \times1 0^{-1} \mathrm{m m}
$$
B. $$
3. 1 2 9 \times1 0^{-1} \mathrm{m m}
$$
C. $$
3. 3 4 0 \times1 0^{-1} \mathrm{m m}
$$
D. $$
1. 7 9 5 \times1 0^{-1} \mathrm{m m}
$$
E. $$
2. 5 7 0 \times1 0^{-1} \mathrm{m m}
$$
F. $$
2. 6 2 2 \times1 0^{-1} \mathrm{m m}
$$
G. $$
1. 8 5 3 \times1 0^{-1} \mathrm{m m}
$$
H. $$
2. 1 5 8 \times1 0^{-1} \mathrm{m m}
$$
I. $$
3. 0 8 1 \times1 0^{-1} \mathrm{m m}
$$
J. $$
1. 9 4 6 \times1 0^{-1} \mathrm{m m}
$$
Answer: A
Design a Chebyshev low-pass filter with the passband cutoff frequency $f_p=3kHz$, maximum passband attenuation $a_{\mathfrak{p}}= 0.2$dB, stopband cutoff frequency $f_s= 12$kHz, and minimum stopband attenuation $a_s= 50$dB. The actual $H_a(s)$ is ().

Options:
A. $H_{a}(s)=\frac{7.2687\times10^{16}}{(s^{2}+8.4748\times10^{3}s+4.2587\times10^{8})(s^{2}+2.0459\times10^{4}s+1.7463\times10^{8})}$
B. $H_{a}(s)=\frac{5.2683\times10^{16}}{(s^{2}+9.4748\times10^{3}s+5.2587\times10^{8})(s^{2}+3.0459\times10^{4}s+2.7463\times10^{8})}$
C. $H_{a}(s)=\frac{8.9972\times10^{16}}{(s^{2}+7.8748\times10^{3}s+3.2587\times10^{8})(s^{2}+1.9459\times10^{4}s+1.5463\times10^{8})}$
D. $H_{a}(s)=\frac{6.2341\times10^{16}}{(s^{2}+8.2748\times10^{3}s+4.0587\times10^{8})(s^{2}+2.1459\times10^{4}s+1.6463\times10^{8})}$
E. $H_{a}(s)=\frac{9.4983\times10^{16}}{(s^{2}+8.5748\times10^{3}s+4.3587\times10^{8})(s^{2}+2.6459\times10^{4}s+1.8463\times10^{8})}$
F. $H_{a}(s)=\frac{4.6571\times10^{16}}{(s^{2}+7.3748\times10^{3}s+2.9587\times10^{8})(s^{2}+1.6459\times10^{4}s+1.2463\times10^{8})}$
G. $H_{a}(s)=\frac{8.7643\times10^{16}}{(s^{2}+7.1748\times10^{3}s+3.8587\times10^{8})(s^{2}+2.3459\times10^{4}s+1.9463\times10^{8})}$
H. $H_{a}(s)=\frac{3.9862\times10^{16}}{(s^{2}+5.4748\times10^{3}s+3.1587\times10^{8})(s^{2}+1.5459\times10^{4}s+1.1463\times10^{8})}$
I. $H_{a}(s)=\frac{6.2892\times10^{16}}{(s^{2}+6.4748\times10^{3}s+3.0587\times10^{8})(s^{2}+1.8459\times10^{4}s+1.3463\times10^{8})}$
J. $H_{a}(s)=\frac{7.6583\times10^{16}}{(s^{2}+6.9748\times10^{3}s+3.6587\times10^{8})(s^{2}+1.7459\times10^{4}s+1.3463\times10^{8})}$
Answer: A
A uniform plane wave with a working frequency of $f = 400 \mathrm{~MHz}$ propagates in the $+z$ direction in a lossless dielectric medium and is incident perpendicularly onto an ideal conductive plane ($z=0$). If the distance between any two adjacent wave nodes in front of the conductive plane is measured to be 12.5 cm, find the relative permittivity of the lossless dielectric medium is ().

Options:
A. 2.25
B. 1.50
C. 0.35
D. 2.00
E. 1.00
F. 0.75
G. 2.50
H. 1.25
I. 0.50
J. 0.25
Answer: J
In water treatment, solid aluminum sulfate must be prepared into a solution of a certain concentration as a coagulant. In a mixing tank for batching, water and solid aluminum sulfate are added to the tank at flow rates of 150 kg/h and 30 kg/h, respectively. After forming the solution, it flows out of the container at a rate of 120 kg/h. Due to thorough mixing, the concentration within the tank is uniform throughout. Initially, the tank contains 100 kg of pure water. Try to calculate the concentration of the solution that flows out of the tank after 1 hour is ().

Options:
A. $$
1 4. 5 1 \%
$$
B. $$
1 1. 4 7 \%
$$
C. $$
1 7. 8 2 \%
$$
D. $$
1 3. 9 9 \%
$$
E. $$
1 6. 3 8 \%
$$
F. $$
1 8. 6 4 \%
$$
G. $$
1 4. 8 6 \%
$$
H. $$
1 5. 2 3 \%
$$
I. $$
1 9. 0 5 \%
$$
J. $$
1 2. 7 5 \%
$$
Answer: H
An aluminum composite pipe with dimensions $\varphi 60 \, \mathrm{mm} \times 3 \, \mathrm{mm}$, and a thermal conductivity of $45 \, \mathrm{W/(m \cdot K)}$, is wrapped with a 30mm thick layer of asbestos and then a 30mm thick layer of cork. The thermal conductivities of asbestos and cork are $0.15 \, \mathrm{W/(m \cdot K)}$ and $0.04 \, \mathrm{W/(m \cdot K)}$ respectively. Question: If the two layers of insulating materials are swapped, assuming the outside temperature of the asbestos remains at 5°C after swapping, what is the cooling loss per meter length of the pipe?

Options:
A. $$
3 4. 5 0 \mathrm{W / m}
$$
B. $$
3 1. 2 5 \mathrm{W / m}
$$
C. $$
3 5. 6 5 \mathrm{W / m}
$$
D. $$
3 3. 1 0 \mathrm{W / m}
$$
E. $$
4 2. 7 0 \mathrm{W / m}
$$
F. $$
3 7. 9 0 \mathrm{W / m}
$$
G. $$
2 9. 8 0 \mathrm{W / m}
$$
H. $$
4 0. 1 5 \mathrm{W / m}
$$
I. $$
3 9. 4 0 \mathrm{W / m}
$$
J. $$
3 6. 0 5 \mathrm{W / m}
$$
Answer: A
A packed tower at atmospheric pressure and 295K is used to absorb ammonia from an ammonia-air mixture using pure water. The mass transfer resistance can be considered to be concentrated in a stationary gas film that is 1 mm thick. At a certain point within the tower, the partial pressure of ammonia is 6.6kPa. The equilibrium partial pressure of ammonia over the water surface can be ignored. Given that the diffusion coefficient of ammonia in air is $0.236 \mathrm{cm}^{2}/\mathrm{s}$, find the mass transfer rate of ammonia at this point.

Options:
A. $$
6. 5 7 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)
$$
B. $$6. 1 8 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
C. $$
3. 5 7 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)
$$
D. $$5. 6 0 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
E. $$7. 0 2 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
F. $$8. 2 3 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
G. $$6. 3 9 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
H. $$4. 9 5 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
I. $$7. 1 5 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
J. $$5. 8 2 {\times} 1 0^{-2} \operatorname{m o l} \! / \! \left( \operatorname{m}^{2} {\cdot} s \right)$$
Answer: A
A container holds oil with a density of $890 \mathrm{kg/m}^{3}$, a viscosity of $0.32 \mathrm{Pa} \cdot \mathrm{s}$, and a depth of $80 \mathrm{cm}$. If small spheres with a density of $2650 \mathrm{kg/m}^{3}$ and a diameter of 5 mm are dropped into the container at intervals of 3 seconds each, and the oil is moving upwards at a speed of $0.05 \mathrm{m/s}$, what is the maximum number of spheres that can be descending simultaneously?

Options:
A. $$ 7 $$
B. $$
11
$$
C. $$ 6 $$
D. $$ 10 $$
E. $$ 13 $$
F. $$ 15 $$
G. $$ 14 $$
H. $$ 12 $$
I. $$ 9 $$
J. $$ 8 $$
Answer: B
Using H-type strong acid cation resin to remove $\mathrm{Na}^{+}$ and $\mathrm{K}^{+}$ from seawater (assuming these are the only two cations present in seawater), given that the concentration of $\mathrm{H}^{+}$ in the resin is $0.3 \mathrm{mol/L}$, and the concentrations of $\mathrm{Na}^{+}$ and $\mathrm{K}^{+}$ in seawater are $0.1 \mathrm{mol/L}$ and $0.02 \mathrm{mol/L}$ respectively, find the concentrations of $\mathrm{Na}$* and $\mathrm{K}$* in the solution at ion-exchange equilibrium is (). It is known that $K_{\mathrm{H^{*}}}^{\mathrm{K^{*}}} = 3.0$ and $K_{\mathrm{H^{*}}}^{\mathrm{Na^{*}}} = 2.0$.

Options:
A. $$
x_{K^{+}}=0. 0 2 0 \;, \; \; x_{N a^{+}}=0. 1 7 2
$$
B. $$
x_{K^{+}}=0. 0 2 6 \;, \; \; x_{N a^{+}}=0. 1 5 5
$$
C. $$
x_{K^{+}}=0. 0 3 0 \;, \; \; x_{N a^{+}}=0. 1 5 0
$$
D. $$
x_{K^{+}}=0. 0 2 4 \;, \; \; x_{N a^{+}}=0. 1 6 3
$$
E. $$
x_{K^{+}}=0. 0 2 3 \;, \; \; x_{N a^{+}}=0. 1 6 2
$$
F. $$
x_{K^{+}}=0. 0 1 9 \;, \; \; x_{N a^{+}}=0. 1 7 0
$$
G. $$
x_{K^{+}}=0. 0 2 5 \;, \; \; x_{N a^{+}}=0. 1 6 0
$$
H. $$
x_{K^{+}}=0. 0 3 1 \;, \; \; x_{N a^{+}}=0. 1 5 8
$$
I. $$
x_{K^{+}}=0. 0 2 8 \;, \; \; x_{N a^{+}}=0. 1 6 1
$$
J. $$
x_{K^{+}}=0. 0 2 2 \;, \; \; x_{N a^{+}}=0. 1 6 8
$$
Answer: E
In the operation of a constant pressure filtration using a specific leaf filter, the resistance of the filter medium can be ignored. At the end of the filtration, $V=0.5 \, \mathrm{m}^{3}$ and $\tau=1 \, \mathrm{h}$. The viscosity of the filtrate is four times that of water. Water is now used for washing, with the wash liquid volume $V_{\mathrm{w}}=0.05 \, \mathrm{m}^{3}$. What is $\tau_{\mathrm{w}}$?

Options:
A. 9min
B. 7min
C. 11min
D. 10min
E. 3min
F. 5min
G. 12min
H. 8min
I. 15min
J. 6min
Answer: E
A certain plate-and-frame filter press has 20 filter frames, each with dimensions of $400 \, \mathrm{mm} \times 400 \, \mathrm{mm} \times 25 \, \mathrm{mm}$. The slurry is a suspension of $\mathrm{CaCO_3}$ at a concentration of 13.9% (by mass). The filter cake has a moisture content of 50% (by mass), and the density of the pure $\mathrm{CaCO_3}$ solid is $2710 \, \mathrm{kg/m^3}$. When the operating pressure difference is 0.3 MPa, the filtration constant is $K = 1.63 \times 10^{-5} \, \mathrm{m^2/s}, \, \, q_{e} = 0.00476 \, \mathrm{m^3/m^2}$. This plate and frame filter press filters the slurry under a pressure difference of 0.3 MPa until the filter cake completely fills the filter frame, and then washes the filter cake with clean water under a pressure of 0.6 MPa, with the amount of wash water being $1/10$ of the filtrate volume. What is the filtration time per cycle?

Options:
A. 158s
B. 175s
C. 177s
D. 180s
E. 165s
F. 172s
G. 162s
H. 166s
I. 160s
J. 169s
Answer: H
Under constant pressure operation of a certain plate-and-frame filter press, after 1 hour of filtration, 2 m³ of filtrate is obtained, with the resistance of the filtration medium being negligible. After 1 hour of filtration, the pressure difference is doubled, and filtration continues for another hour. Given that the compressibility index of the filter cake is $s=0.2$, how much more filtrate is obtained?

Options:
A. $1.234m^3$
B. $1.522m^3$
C. $1.927m^3$
D. $1.580m^3$
E. $1.643m^3$
F. $1.455m^3$
G. $1.702m^3$
H. $1.789m^3$
I. $1.309m^3$
J. $1.618m^3$
Answer: I
For constant pressure filtration, when the filtration area is doubled, by how many times does the filtration rate $\mathrm{d}V/\mathrm{d}\tau$ increase? (The resistance of the filtration medium can be ignored.)

Options:
A. 5
B. 2
C. 4
D. 6
E. 8
F. 3.5
G. 3
H. 7
I. 1.5
J. 0.5
Answer: C
A certain dust chamber is 2 meters high, 2 meters long, and 3 meters wide, and is used to remove dust particles from dust-laden gases. Under operating conditions, the density of the spherical dust particles is 2000 kg/m³, and the density of the dust-laden gas is 1.8 kg/m³ with a viscosity of 0.025 mPa·s. If the requirement is to remove all dust particles larger than 25 μm, what is the dust-laden gas processing capacity of this dust chamber?

Options:
A. $589. 0m^3/h$
B. $595. 8m^3/h$
C. $583. 4m^3/h$
D. $588. 2m^3/h$
E. $580. 2m^3/h$
F. $592. 7m^3/h$
G. $587. 5m^3/h$
H. $585. 6m^3/h$
I. $600. 0m^3/h$
J. $590. 3m^3/h$
Answer: G
Using a jacketed heat exchanger with a stirrer, an organic substance with a mass of 750 kg and a specific heat capacity of 2.2 kJ/(kg°C) is cooled from 70°C to 40°C. It is known that the overall heat transfer coefficient of the jacketed heat exchanger is K = 160 W/(m²°C), with a heat transfer area of 3 m². The coolant water flow rate is 0.1 kg/s, with an inlet temperature of t₁ = 20°C. Ignoring heat loss, and assuming uniform temperature within the stirred liquid, how long is required to complete the cooling process?

Options:
A. 7.85h
B. 5.62h
C. 2.58h
D. 4.27h
E. 8.09h
F. 6.13h
G. 0.89h
H. 0.34h
I. 1.47h
J. 3.15h
Answer: I
In a certain factory, the height of the packing layer in the absorption tower is 3 meters. A pure solvent is used for countercurrent isothermal absorption to remove harmful components from the tail gas. The concentration of harmful components in the inlet gas is 0.04 (mole fraction, the same below), the concentration of harmful components in the outlet gas is 0.008, and the concentration of harmful components in the outlet liquid is 0.03. It is known that within the operating range, the phase equilibrium relationship is $y = 0.8x$. What is the overall gas phase mass transfer unit height, $H_{\mathrm{OG}}$, of the packed tower?

Options:
A. 1.082m
B. 1.254m
C. 0.920m
D. 1.500m
E. 1.070m
F. 1.940m
G. 1.600m
H. 1.860m
I. 0.500m
J. 2.064m
Answer: A
To determine the height of the mass transfer unit $H_{\mathrm{OG}}$ for novel packing, use the desorption of $\mathrm{CO_{2}}$ aqueous solution. In the experimental tower, the height of the packing layer is 2.5m, with a water flow rate into the top of the tower of $4000 \, \mathrm{kg}$ water/h. The concentration of $\mathrm{CO_{2}}$ is $8 \times 10^{-5}$ (mole fraction). Fresh air containing no $\mathrm{CO_{2}}$ is introduced at the bottom of the tower at a rate of $6 \, \mathrm{kg} / \mathrm{h}$, and the liquid concentration in the tower is measured as $4 \times 10^{-6}$ (mole fraction). The phase equilibrium relation is given by $y = 1420 x$ (mole fraction). What is the gas concentration (mole fraction) in the tower?

Options:
A. $$
y_{out}=0.0838
$$
B. $$
y_{out}=0.0821
$$
C. $$
y_{out}=0.0702
$$
D. $$
y_{out}=0.0750
$$
E. $$
y_{out}=0.0769
$$
F. $$
y_{out}=0.0784
$$
G. $$
y_{out}=0.0912
$$
H. $$
y_{out}=0.0875
$$
I. $$
y_{out}=0.0953
$$
J. $$
y_{out}=0.0816
$$
Answer: J
26 years old, pregnant: missed period for 8 weeks, diagnosed with early pregnancy, requests abortion. Examination: uterus size corresponds to 50 days of pregnancy, right fallopian tube is thickened with obvious tenderness, white blood cell count is 11.6 x 10⁹/L, neutrophils are 0.80. Which of the following treatments is most appropriate?

Options:
A. Medication To Induce Labor Naturally
B. Endometrial Biopsy For Histological Examination
C. Surgical Removal Of Ovarian Cyst
D. Observation With Weekly Blood Counts
E. After anti-inflammatory treatment, the uterus is aspirated to cause miscarriage
F. Hormonal Therapy For Tubal Dysfunction
G. Immediate Laparoscopic Surgery For Ectopic Pregnancy
H. Intravenous Antibiotics And Observation
I. Laparotomy For Pelvic Abscess Drainage
J. Oral Antibiotics With Bed Rest
Answer: E
A 24-year-old pregnant woman. At 36 weeks of G1P0 intrauterine pregnancy, she has experienced lower limb edema for 8 months, generalized edema for the past week, dizziness, headache, and blurred vision in the last 24 hours. She had a sudden seizure 3 hours ago lasting 1 minute and has had two recurrent episodes before arriving at the hospital. Upon examination in the hospital, her blood pressure was 180/120 mmHg, heart rate 100/min, respiratory rate 30/min, and she was in a comatose state. Heart and lungs were unremarkable. The fundus height was 35 cm above the pubic symphysis, fetal heart rate was 110/min, and a vaginal examination showed cervical dilation of 1.5 cm. What obstetric management should be employed?

Options:
A. Administer methotrexate for ectopic pregnancy emergency procedure
B. Apply fundal pressure to accelerate the progression of labor
C. Post-dural puncture headache after spinal anesthesia cesarean section
D. Schedule a non-stress test for fetal well-being assessment
E. Prepare for a forceps delivery to assist during second-stage labor
F. Utilize cervical cerclage for incompetent cervix stabilization
G. Initiate bed rest and hydration for mild preeclampsia treatment
H. Conduct a uterine compression suture for postpartum hemorrhage
I. Perform a nitrazine test to confirm premature rupture of membranes
J. Recommend fetal surgery for congenital diaphragmatic hernia repair
Answer: C
Calculate the heat of combustion of propane at 600 K and 101,325 Pa, assuming that the water in the combustion products is in the gaseous phase.

Options:
A. $$
- 2 \ 3 8 8 \ 1 5 9. 4 \ \mathrm{J / m o l}
$$
B. $$
- 2 \ 6 1 0 \ 6 4 3. 8 \ \mathrm{J / m o l}
$$
C. $$
- 2 \ 2 5 3 \ 6 7 4. 0 \ \mathrm{J / m o l}
$$
D. $$
- 2 \ 3 4 5 \ 7 5 4. 9 \ \mathrm{J / m o l}
$$
E. $$
- 2 \ 1 3 4 \ 8 9 5. 5 \ \mathrm{J / m o l}
$$
F. $$
- 2 \ 4 0 0 \ 5 2 0. 3 \ \mathrm{J / m o l}
$$
G. $$
- 2 \ 3 1 9 \ 7 0 8. 7 \ \mathrm{J / m o l}
$$
H. $$
- 2 \ 2 7 0 \ 9 8 2. 6 \ \mathrm{J / m o l}
$$
I. $$
- 2 \ 3 1 5 \ 7 6 8. 1 \ \mathrm{J / m o l}
$$
J. $$
- 2 \ 4 5 9 \ 3 4 1. 2 \ \mathrm{J / m o l}
$$
Answer: G
The concentration of methyl methacrylate $( \mathbf{M}_{1} )$ is 5 mol· $\mathbf{L}^{-1}$, and the concentration of 5-ethyl-2-vinylpyridine is 1 mol· $\mathbf{L}^{-1}$. The reactivity ratios are $r_{1}=\! 0.40, r_{2} \!=\! 0.69$. What is the molar ratio of the two monomers when the composition of the copolymer is the same as the monomer composition?

Options:
A. $\frac{4}{9}$
B. $\frac{21}{34}$
C. $\frac{8}{15}$
D. $\frac{19}{35}$
E. $\frac{25}{41}$
F. $\frac{17}{33}$
G. $\frac{12}{23}$
H. $\frac{5}{7}$
I. $\frac{6}{13}$
J. $\frac{29}{56}$
Answer: F
The conformation of amines with a nitrogen atom as the chiral center depends on the inversion barrier of the lone pair of electrons on the nitrogen atom, which is ____ (25 kJ·mol⁻¹). In comparison, the inversion barrier of PH is ____ , allowing enantiomers to interconvert rapidly at room temperature. As a result, their isomers cannot be separated.

Options:
A. Unstable, too small, $160 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
B. Stable, too large, $100 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
C. Stable, too small, $180 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
D. Stable, too large, $150 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
E. Stable, too small, $200 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
F. Unstable, too small, $140 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
G. Unstable, too large, $120 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
H. Unstable, too large, $110 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
I. Unstable, too large, $170 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
J. Stable, too small, $130 \mathrm{~kJ ~} \cdot\mathrm{mol}^{-1}$
Answer: F
Feathered shank (F) in chickens is dominant over bare shank, while the pea comb gene (P) is dominant over the single comb gene. When a feathered shank, pea comb rooster mates with a bare shank, pea comb hen, all offspring have feathered shanks. Most of them have pea combs, but some have single combs. What are the genotypes of the parents?

Options:
A. $$
Ffpp \times ffPP
$$
B. $$
FfPp \times FfPP
$$
C. $$
FFpp \times ffPp
$$
D. $$
ffPp \times FFPp
$$
E. $$
FfPP \times FFpp
$$
F. $$
FFPp \times ffPP
$$
G. $$
Ffpp \times ffPp
$$
H. $$
FFpp \times ffPP
$$
I. $$
FFPP \times ffpp
$$
J. $$
FfPP \times ffPp
$$
Answer: F
An object moves upward along a circular path. To maintain a constant horizontal speed of $v=200 \text{ mm/s}$, find the velocity and acceleration of the object when $\theta=0$ and $60^\circ$.

Options:
A. $350mm/s$, $250mm/s^{2}$.
$300mm/s$, $300mm/s^{2}$.
B. $500mm/s$, $600mm/s^{2}$.
$100mm/s$, $400mm/s^{2}$
C. $600mm/s$, $300mm/s^{2}$.
$200mm/s$, $500mm/s^{2}$.
D. $200mm/s$,$200mm/s^{2}$
$400mm/s$, $800mm/s^{2}$
E. $250mm/s$, $150mm/s^{2}$.
$300mm/s$, $300mm/s^{2}$.
F. $100mm/s$, $400mm/s^{2}$.
$450mm/s$, $700mm/s^{2}$
G. $150mm/s$, $100mm/s^{2}$.
$400mm/s$, $800mm/s^{2}$
H. $450mm/s$, $700mm/s^{2}$.
$200mm/s$, $500mm/s^{2}$
I. $300mm/s$, $300mm/s^{2}$.
$400mm/s$, $800mm/s^{2}$
J. $200mm/s$, $500mm/s^{2}$.
$600mm/s$, $300mm/s^{2}$
Answer: D
A beam with a length of $12 ft$ is subjected to a uniformly distributed load of $100 lb/ft$ over a section starting $4 ft$ from the left end. The load increases uniformly from this $100 lb/ft$ to a maximum of $300 lb/ft$ at the right end. What are the reactions at the ends of the beam?

Options:
A. $$
{R_L}=780lb
$$
$$
{R_R}=1220lb
$$
B. $$
{R_L}=790lb
$$
$$
{R_R}=1210lb
$$
C. $$
{R_L}=770lb
$$
$$
{R_R}=1230lb
$$
D. $$
{R_L}=740lb
$$
$$
{R_R}=1260lb
$$
E. $$
{R_L}=810lb
$$
$$
{R_R}=1190lb
$$
F. $$
{R_L}=820lb
$$
$$
{R_R}=1180lb
$$
G. $$
{R_L}=760lb
$$
$$
{R_R}=1240lb
$$
H. $$
{R_L}=730lb
$$
$$
{R_R}=1270lb
$$
I. $$
{R_L}=800lb
$$
$$
{R_R}=1200lb
$$
J. $$
{R_L}=750lb
$$
$$
{R_R}=1250lb
$$
Answer: A
For a certain ideal gas, the initial state is $p_{\mathrm{1}}=520 \mathrm{kPa},\ V_{\mathrm{1}}=0.1419 \mathrm{m}^3$. After an exothermic expansion process, the final state is $p_{2}=170 \mathrm{kPa},\ V_{2}=0.2744 \mathrm{m}^3$. The enthalpy change for the process is $\Delta H=-67.95 \mathrm{kJ}$. Given that the constant pressure specific heat capacity of the gas is $c_{p}=5.20 \mathrm{kJ}/(\mathrm{kg} \cdot \mathrm{K})$, which is a constant, find the specific heat capacity at constant volume and the gas constant $R_{g}$.

Options:
A. $$
2.500 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
B. $$
1.850 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
C. $$
2.077 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
D. $$
1.750 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
E. $$
2.475 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
F. $$
3.125 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
G. $$
2.300 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
H. $$
2.600 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
I. $$
1.980 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
J. $$
1.625 \mathrm{kJ} / ( \mathrm{kg} \cdot\mathrm{K})
$$
Answer: C
In a plug flow reactor, an acetic acid hydrolysis reaction is carried out. The concentration of acetic acid entering the reactor is $c_A = 0.3 \text{ kmol} \cdot m^{-3}$, and the required conversion rate is $x_A = 70\%$ (based on the feed). The feed volumetric flow rate is $q_v = 0.02 m^3 \cdot \text{min}^{-1}$. If a large amount of water is added, the acetic acid hydrolysis reaction can be considered a first-order reaction, expressed as $-r_A = k c_A$, where $k = 0.38 \text{ min}^{-1}$. Determine:
What should the reactor volume be to achieve the above conversion rate using a plug flow reactor?

Options:
A. $130L$
B. $200L$
C. $160L$
D. $123L$
E. $150L$
F. $190L$
G. $140L$
H. $165L$
I. $175L$
J. $180L$
Answer: D
Using a 300mm diameter packed column with a packing height of 3.2m, absorb a certain harmful gas from the air with clean water. Known data: the gas-phase inlet and outlet concentrations are $Y_1=0.1, Y_2=0.004$, air flow rate $V=10 \text{ kmol} \cdot \text{h}^{-1}$, water flow rate $L=20 \text{ kmol} \cdot \text{h}^{-1}$, and under the operating conditions, the system follows Henry's law $Y^*=0.5X$. What is the gas-phase outlet concentration when the packing height is increased by 2m?

Options:
A. 0.00090
B. 0.00072
C. 0.00095
D. 0.00063
E. 0.00078
F. 0.00085
G. 0.00110
H. 0.00105
I. 0.00080
J. 0.00098
Answer: D
Under normal pressure, harmful components in the air are absorbed with water in the absorption tower using a co-current method. At 20°C, measurements at the bottom of the tower show $x=0.008$ (mole fraction, the same below), and $y=0.04$. If the equilibrium relationship follows Henry's law, with a solubility coefficient $H=1.8 \, \text{kmol}\cdot\text{m}^{-3}\cdot\text{kPa}^{-1}$, gas-phase mass transfer coefficient $k_{G}=4.4\times10^{-5} \, \text{kmol}\cdot\text{m}^{-2}\cdot\text{s}^{-1}\cdot\text{kPa}^{-1}$, and liquid-phase mass transfer coefficient $k_{L}=8.2\times10^{-4} \, \text{m}\cdot\text{s}^{-1}$. What is the mass transfer rate $N_{\Lambda}$ at the bottom of this tower?

Options:
A. $9.402 \times 10^{-7} \, \text{mol} \cdot \text{m}^{-2} \cdot \text{min}^{-1}$
B. $4.567 \times 10^{-4} \, \text{mol} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
C. $3.150 \times 10^{-6} \, \text{kg} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
D. $2.431 \times 10^{-5} \, \text{kmol} \cdot \text{m}^{-2} \cdot \text{hr}^{-1}$
E. $7.540 \times 10^{-5} \, \text{mol} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
F. $1.697 \times 10^{-4} \, \text{kmol} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
G. $8.765 \times 10^{-6} \, \text{kmol} \cdot \text{m}^{-2} \cdot \text{hr}^{-1}$
H. $5.672 \times 10^{-6} \, \text{kmol} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
I. $2.985 \times 10^{-5} \, \text{kmol} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
J. $1.230 \times 10^{-3} \, \text{kg} \cdot \text{m}^{-2} \cdot \text{s}^{-1}$
Answer: F