Hard to visualize GC question

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

SonhosDaVida

Full Member
7+ Year Member
Joined
May 23, 2014
Messages
265
Reaction score
37
I am having a hard time visualizing where normal expiration of 1800 mL falls on the cycle of breathing diagram. I know tidal volume is the normal amount of air displaced during expiration and inhalation, so where would normal expiration fall in the graph? Directly below 1200 mL? Wouldn't that be the expiratory reserve volume?

I'm having a hard time visualizing this problem.

So is inspiration volume the same as expiration volume?

It is Q5.

Members don't see this ad.
 

Attachments

  • 15320275953592310062309843937851.jpg
    15320275953592310062309843937851.jpg
    123.4 KB · Views: 53
  • 15320276521944798020697567821849.jpg
    15320276521944798020697567821849.jpg
    186.6 KB · Views: 48
Hi, SonhosDaVida--

No need to apologize--it is no bother.

Figure 1 shows the breathing capacities of a 93.7 kg male. Question 5 is not dealing with a 93.7 kg male because the numbers don't seem to correlate. Once that dawned on me, it made a little more sense. For question 5, the analogy would be like the first two breaths in figure 1 but with different scale on the y-coordinate; the troughs would be at a height of 1800 mL (the volume of the lungs after normal expiration). The distance from the trough to the crest of normal inhalation (tidal volume) would be 1200 mL. Therefore, the crest would be at a height of 3000 mL (1800 mL trough + 1200 mL takes the crest to 3000 mL).

Use VP=nRT; we know that nRT are constant.
Therefore: V1P1 = V2P2.
You can set V1 = 1800 mL and P1 = 1 atm (you can use any number, but 1 is easiest) and V2 = 3000 mL:
(1800)(1) = (3000)(x)
1800/3000 = x
3/5 = x
0.60 = x
So the pressure went from 1 to 0.60; a decrease of 40%.

I hope that helps.
 
  • Like
Reactions: 1 user
Top