# Tutor profile: Nikko D.

## Questions

### Subject: MATLAB

Use MATLAB to construct an anonymous function $$f$$ which can take variables V (volume), T (temperature) and van der Waals constants $$n$$, $$a$$, and $$b$$ from the van der Waals equation: $$(p + \frac{n^2a}{V^2})(V-nb) = nRT$$ Then, make a script function called “vanderwaals” which should have the following form: function [P, V, T] = vanderwaals ($$f$$, T1, T2, nT, V1, V2, nV, varargin) This "vanderwaals" function should generate temperature within the limits T1 to T2 with step of nT = 10 units (fixed). Also, this function should generate volume values between limits V1 to V2 with step of nV, and $$f$$ is the anonymous function created earlier. Your "vanderwaals" function should return Pressure (P), Volume (V) and Temperature (T) of the non-ideal fluid. Also, it should have two plots: (1) a mesh contour plot between T, V, and P (2) a plot between V and P for ten values of temperature (isotherms). Use the following inputs to produce a sample output: T1 = 100 K T2 = 800 K nT = 10 K V1 = 50 $$cm^3$$ V2 = 200 $$cm^3$$ nV = 150 $$cm^3$$ $$n$$ = 1 mole $$a$$ = 0.5537 $$ \frac{J*m^3}{mol^2}$$ $$b$$ = 30.4 $$ \frac{cm^3}{mol}$$

%% MATLAB solution clc clear all close all R = 8.3144621; %searching for appropriate R value, MPa, K, cm3, mol = 8.314 4621(75) [https://www.cpp.edu/~lllee/gasconstant.pdf] f = @(V,T,a,b,n) n.*R.*T./(V-n.*b) - (n.^2).*a./(V.^2); % anonymous function "f" which solves for P [P, V, T] = vanderwaals(f, 100, 800, 10, 50, 200, 5); %running script for the sample input %% Script function "vanderwaals" function [P, V, T] = vanderwaals(f, T1, T2, nT, V1, V2, nV, varargin) a = 0.5537.*(100^3); %conversion of m3 to cm3, and J*m to MPa b = 30.4; n = 1; % mole V = V1:nV:V2; %vector for volume T = T1:nT:T2; %vector for temperature %% PVT mesh and plot vcount = numel(V); %number of V elements tcount = numel(T); %number of T elements i = 1; %dummy counter j=1; %dummy counter P = []; %mesh base for i = 1:tcount for j = 1:vcount P(i,j) = f(V(j),T(i),a,b,n); %creation of P mesh end end %%Plotting %For contour mesh plot figure() meshc(T,V, P') xlabel('Temperature(K)') ylabel('Volume(cm^3)') zlabel('Pressure(MPa)') %For stacked PV plot figure() i=1; %dummy counter for i = 1:round(tcount./10):tcount % use round(tcount./10) to restrict output to ten isotherms only plot(V,P(i,:)) hold on end xlabel('Volume(cm^3)') ylabel('Pressure(MPa)') V = V'; T = T'; end

### Subject: Chemistry

For every ten meters of descent under the ocean, the pressure increases by around 1 atmosphere. At depths of 1,000 meters, the pressure will be around 100 times the pressure at sea level. Most fish have a gas-filled organ called the swim bladder which helps them maintain buoyancy. Fishes at so-called twilight zones or 1,000 meters below sea level have adapted to pressures at this depth. If a fisher rapidly brings a fish from this region to the surface, the swim bladder of the fish expands so much that it can crush its vital organs, causing the fish to die. Which gas law explains this? A. Boyle’s law B. Charles’ law C. Graham’s law D. Gay-Lussac’s law

The problem described changes between two main variables for the phenomenon observed: (1) Pressure - comparison with the sea level pressure using depth as multiplier (100 times more at 1000 m below) (2) Volume - indicated in the expansion of the swim bladder, which was also described as gas-filled. Given these two variables, the gas law that explains the relationship between these is Boyle's Law: $$P_1V_1 = P_2V_2$$ $$where$$ $$Point \ 1 \ is \ at \ 1000 \ m \ below \ sea \ level $$ $$Point \ 2 \ is \ at \ sea \ level $$ The phenomenon was observed after $$rapidly$$ bringing the fish from a higher pressure point ( around 100 times sea level pressure) to a lower pressure point ( sea level pressure). Let $$P_1 = 100P_2$$ Substituting values: $$100P_2V_1 = P_2V_2$$ Simplifying: $$100V_1 = V_2$$ Thus, the volume of the swim bladder rose to 100 times its original volume and crushes the vital organs of the fish during its expansion.

### Subject: Physics

Airbags are installed in cars to diminish the injury to a passenger during a car crash. Which primarily explains this benefit? A. The airbag reduces the force by reducing the passenger’s velocity change. B. The airbag reduces the impulse by reducing the change in the passenger’s momentum. C. The airbag reduces the force by increasing the time it takes the passenger to stop moving. D. The airbag reduces the change in velocity by reducing the time it takes the passenger to stop moving.

Impulse equals force times the time over which it acts: $$Impulse = F \Delta t$$ Also, according to the impulse-momentum equation, the impulse of an object acted upon by a force $$F$$ over time $$\Delta t$$ is equal to the change in the object’s momentum $$m \Delta v$$ where $$m$$ is the object’s mass and $$\Delta v$$ is the change in the object’s velocity. $$Impulse = F \Delta t = m \Delta v$$ During a sudden stop, the final velocity must go to zero so $$\Delta v = initial\ velocity $$ and is a constant value. Moreover, the passenger’s mass, $$m$$, is also a constant value. However, the force $$F$$ acting on the passenger’s mass and the time duration it takes for the passenger to stop moving, $$\Delta t$$, can change as long as the impulse is constant. In effect, the increase in contact time, $$\Delta t$$, brought by the airbag during a crash results in a decrease in the force applied to the passenger, $$F$$, due to their inverse relationship. Therefore, the correct answer is letter C.