# Tutor profile: Sam C.

## Questions

### Subject: Chemistry

A tank with a partition (or separator) is holding 1 mole of H2(g) and 2 moles of N2(g) at 1 atm. The partition is removed and the gas begins to mix in the tank. At equilibrium, what is the pressure of the system due to the H2(g)? assume ideal conditions.

1) As always, I would suggest by diagraming this system and labeling its components. Make a box and draw a line through the middle. In the left half you have a mole of H2(g) (use an X to represent this gas), and on the right, N2(g) (use an O to represent this gas). The pressure on both sides is 1 atm. Next to this picture, draw a new box with all of the X's and O's mixed within the system. 2) We know that we are looking for the partial pressure of the hydrogen gas in the heterogeneous (or "not-same" system). To do so requires a lot more semantics than calculations. a) If the pressure of the system was 1 atm before the divider was removed, what is the total pressure of the system after the divider is taken away? Well, the container hasn't changed its volume, temperature, or total moles but this is an ideal gas so if Pv=nRT then P=nRT/v. Since n, T, and V don't change, pressure won't either! Now we can say that the total pressure is equal to the pressure of the hydrogen gas plus the pressure of the nitrogen gas. The pressure associated with each gas will be referred to as a partial pressure. b) We can describe the total pressure, P, in terms of the partial pressures, pp, with the equation P=pp(H2)+pp(N2). Furthermore, we can say the the pressure of each gas is proportionate to the relative amount of each gas in the container. Think about it this way, if you have three eggs and one apple in a refrigerator, the pressure of the system would be expressed as 75% from the eggs and 25% from the apple because of the 4 items 3/4 are eggs and 1/4 are apples. When it comes to gases, we can figure out the ratio of the amounts of gases by taking the amount of one gas and dividing it by the total amount of gasses. To keep our units similar, we will make sure everything is in moles. Not so surprisingly we call this the molar ratio of a gas. c) Now that we understand the contributing factors of this problem we can simply plug in our information into the relationships we have identified. Since there is 1 mol of H2 gas and 2 moles of N2 gas, we know that there are 3 total moles of gas. 1/3 or 33% of the system's gas is thusly hydrogen molecules. The total pressure of the system is 1 atm and because the system is 1/3 H2, we can say that hydrogen contributes 1/3 of the total 1 atm of pressure. d) pp(H2)=(1 mole H2 gas/ 3 total moles gas)x(1 atm total pressure) = 0.33 atm of pressure.

### Subject: Calculus

f(x)=x^3+ln(x)-sin(x) find f'(x)

To find f'(x) we will break our function down into the three sub derivatives. 1) we can use the power rule to derive x^3 such that x^n derived is n*x^(n-1). This works out neatly to be 3x^2 2) ln(x) is a derivative I suggest you memorize although the proof isn't too hard when considering that this notation is just an exponential function e^y=x. ultimately we find that ln(x) derives as 1/x*x/dx (remembering to use the chain rule). of course x/dx is just 1 so our derivative is 1/x. 3) -sin(x) is another derivative to memorize but it is simply - cos(x) 4) to put it all together we get that f'(x) = 3x^2+(1/x)-cos(x)

### Subject: Physics

You pick up your 1kg Physics textbook and walk to school 2km away. By the time you get there you feel exhausted. How much work did you do to lug that book all the way to school?

The mechanics of this problem are not difficult but you may notice some tricky steps along the way. 1) work is simply a force applied over a distance: W=F*d. This is where we must apply our first trick. Force is an accelerated mass: F=m*a. We know that the book weighs 1kg but how is it being accelerated independently of us touching it? That's right, because you are standing on planet earth, we experience a constant acceleration of gravity, g. The value for this acceleration is 9.81 (m/s^2). Now, if we want to calculate our force, we don't even need to pull out our calculator! F=m*a = 1kg*9.81(m/s^2) but anything times 1 is just itself so F= 9.81 kg*m/s^2. Since we applied that force of 9.81 newtons over a distance of 2km we can just plug and chug into our work equation. W=F*d = 9.81(kg*m/s^2)*2km BUT WAIT we must look at the units. Since our force include meters in its unit we need to also make sure that our distance is in meters but that shouldn't be too hard. 2km is 2000m so we can finally calculate that W=9.81(kg*m/s^2)*2000m = 19620 (kg*m^2/s^2). The final step is to clean up our answer by using sig. figs. In the original problem, I gave you two numbers with only one significant figure each. That means our answer should only be one significant figure as well. 19620 can be written in scientific notation as 1.9620*10^4 and when take only one sig fig we get that our answer will be rounded to 2*10^4 (kg*m^2/s^2). If you are really good with your units you will recognize that this is also the same as 2*10^4 J or 20 kJ.

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