Can someone explain in detail how to do this problem? Thereare two part to it. A
ID: 681609 • Letter: C
Question
Can someone explain in detail how to do this problem? Thereare two part to it. At 1400 mm Hg and 286 K a diver exhales a 208 mL bubble of airthat is 77% N2 , 17% O2, and 6%CO2 by volume. How many milliliters would the volume of the bubble be at thesurface and 298 K and how many moles of N2 are in thebubble? Can someone explain in detail how to do this problem? Thereare two part to it. At 1400 mm Hg and 286 K a diver exhales a 208 mL bubble of airthat is 77% N2 , 17% O2, and 6%CO2 by volume. How many milliliters would the volume of the bubble be at thesurface and 298 K and how many moles of N2 are in thebubble?Explanation / Answer
TheIdeal Gas Law says PV = nRT. Here, we keep n and R constant, soPV/T must be kept constant as well. Initially, PV/T = 1400mmHg *208mL / 286K. This must equal the new P at the surface, which is760mmHg (1 atm), so 1400mmHg * 208mL / 286K = 760mmHg * V / 298K.This gives a value V of 399mL. Now, remember that at 1 atm, a moleof gas always takes up 22.4L. So, dividing we have .339L / 22.4L =.015 total moles of gas. 77% of that is nitrogen, so .015 * .77 =.01155 moles of nitrogen.