Question 51

[jnguyen07080]

Hi there,

For question number 51, I don't understand why you use the equation work = pressure * change in volume. Since the question states that this is an isothermal process, doesn't that mean that the pressure varies along with the change in volume? If that is the case wouldn't you have to multiply the change in pressure by the change in volume to figure out the amount of work done?

Thanks

[admin]

The passage states (final paragraph) that even though the height might change (indicating a change in volume), the pressure remains unchanged at 3 atm.

[hfm03]

The value that was used for A was 50cm^2. However, don't you suppose to change it m^2 which is going to give us about .005m^2.
(.005*.1-.005*.005)3*1.01*10^5=142J

would you explain that for me please?

[jvuofm1709]

Hmmmm. Although the passage states that the pressure remains constant at 3 atm. This problem states that the expansion is ISOTHERMAL, that changes the game. When the expansion was NOT isothermal, we could change Volume and Temperature. That means V/T=nR/P. nR/P was a constant, so a change in volume HAD to lead to a change in temperature. This is the hallmark of a IRREVERSIBLE expansion, and the work performed is calculated by= P*deltaVolume.

Now, when the question says that the expansion is isothermal, that changes everything. Isothermal, literally means that the temperature does not change, it is constant. so now pv=nRT. nRT is our new constant, as volume changes...pressure has to change...its inherent to the circumstances. You cant change the volume of this gas at constant temperature and keep the pressure constant. This isothermal circumstance is a REVERSIBLE expansion, and work is calculated by work=-nRT*ln(Vfinal/Vinitial) or work=-nRT*ln(Pinitial/Pfinal). Seeing as how these equations are derived from calculus based integrals sums i would hope this would be a test question.

[jsfkt78927]

isothermal means constant temp. ---> even at a constant temp you can have an increase in volume or pressure.

ie think about constant velocity -->does this mean you'd be standing still? Not necessarily.

The units are stupid...

The conversion you're thinking of is very timely,

Ie:

50 cm^2 (1m/100cm) = 0.5 cm^2 (0.05cm) = 0.025 cm^3

----> that's right, right?? No

square the conversion factor to match the area

Ie:

50 cm^2 (1m/100cm)^2 => 50cm^2 (1m^2/1000cm^2) = 0.050 m^2 -----> is that right? Yeah

But, here's the caveat ---- you dont need to do that!

GS fails to mention this (so do their tests IMO), but 1L = 1000 cm^3

Ie:

50 cm^2 x 5 cm = 250 cm^3 (1L/1000 cm^3) = 0.25 L

Unfortunately, if you jumped through all those hoops of mathematical fire (which getting a question like this on the real MCAT is 1/1000000 IMO -- which is why i hate GS b/c of the unpractical type of practice) you still need to convert to m^3.

Who knows that conversion off the top of their heads???

Well 1 m^3 = 1000L

So you get 0.25L (1m^3/1000L) = 0.00025 m^3

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Note you could have done this by converting the area given to meters and then the x1 and x2 to meters initially....