Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
All five of these are the same equation, so it is worth writing it down once and then not thinking about it again. Everything below is the combined gas law with the constant quantity cancelled off, and two rules make the whole set mechanical: temperature goes in kelvin every single time, and whatever is held constant simply drops out of both sides.
P1 V1 / T1 = P2 V2 / T2 (combined gas law) K = C + 273.15
6) Nitrogen at constant volume. V cancels, so P/T is what stays fixed. T1 = 127 + 273.15 = 400.15 K and T2 = 274 + 273.15 = 547.15 K.
P2 = P1 x (T2 / T1) P2 = 2 atm x (547.15 / 400.15) P2 = 2 atm x 1.3674 = 2.73 atm
If your book rounds to 273 instead of 273.15 you get 2.735, which reports as 2.74 atm. Pick one convention and stay with it. Worth noticing: going from 127 C to 274 C looks like doubling the temperature, and the pressure does not double, because 274 C is not twice as hot as 127 C on an absolute scale. It is 1.37 times as hot.
7) Both pressure and temperature change. Nothing cancels, so use the full law and solve for V2. T1 = 43 + 273.15 = 316.15 K and T2 = 14 + 273.15 = 287.15 K.
V2 = V1 x (P1 / P2) x (T2 / T1) V2 = 11 mL x (0.849 / 0.997) x (287.15 / 316.15) V2 = 11 mL x 0.8516 x 0.9083 V2 = 8.51 mL
Both effects push the same direction here, since the gas is squeezed to a higher pressure and cooled at the same time, so the answer has to land below 11 mL. If you got a number above 11, one of the two ratios went in upside down.
8) STP, then compressed and heated. STP supplies the starting conditions: P1 = 101325 Pa and T1 = 273.15 K. T2 = 35 + 273.15 = 308.15 K.
P2 = P1 x (V1 / V2) x (T2 / T1) P2 = 101325 Pa x (619 / 228) x (308.15 / 273.15) P2 = 101325 Pa x 2.7149 x 1.1281 P2 = 3.10 x 10^5 Pa (about 3.06 atm)
One caution, because it is the only real trap in this problem. The older definition of STP, and the one this question is built on, is 0 C with 1 atm = 101325 Pa. IUPAC now defines it as 0 C with exactly 100 kPa, which would give 3.06 x 10^5 Pa instead. The two answers differ by about 1.3 percent, so check which definition your course uses before you submit.
9) The tire, constant volume, solved backwards. You have both pressures and you want the temperature, so rearrange P/T = constant for T2.
T2 = T1 x (P2 / P1) T2 = 271 K x (1.85 / 2.8) T2 = 271 K x 0.6607 = 179.05 K C = 179.05 - 273.15 = -94.1 C
That is the answer the assignment is looking for, and it is also physically absurd, because no stretch of road between New York and Boston sits at 94 degrees below zero. The reason is that the problem pours gauge pressures into a law that only accepts absolute pressures. A tire gauge reads zero at atmospheric, so a gauge reading of 2.8 atm is really 3.8 atm absolute. Rerun it that way and T2 = 271 x (2.85 / 3.8) = 203 K, or about -70 C. Still far too cold, because a pressure drop of a third is much larger than any real drive produces. Submit -94 C, but know why no tire behaves like that.
10) The aerosol can. Identical shape to number 9, running the other way. T1 = 29 + 273.15 = 302.15 K.
T2 = T1 x (P2 / P1) T2 = 302.15 K x (4.74 / 4.5) T2 = 302.15 K x 1.0533 = 318.26 K C = 318.26 - 273.15 = 45.1 C
45 C is about 113 F, which is a plausible hot beach, so this one passes the sanity check that number 9 fails. It is also exactly why aerosol cans carry a warning not to store them above 50 C. The pressure climbs with absolute temperature whether the can is built for it or not.
Two habits will carry you through the rest of this set faster than memorizing separate laws. Do not keep Boyle, Charles and Gay-Lussac as three formulas; write the combined law and cross off whatever the problem holds constant. And before computing, say out loud whether the answer should come out larger or smaller than what you started with. Every one of these five can be checked that way in about five seconds, and it catches an inverted ratio, which is essentially the only mistake this kind of problem produces.
