TBR Lecture 6 question #40

[SirWilliamOsler]

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TBR Chemistry Lecture 6

#40. For lungs with a tidal volume of 400 mL and a total volume following normal expiration of 1200 mL, how does internal pressure change to cause inspiration?

A. It increases by 33%
B. It increases by 25%
C. It decreases by 25%
D. It decreases by 33%


I have been having some trouble with percentage problems. I cannot understand how the answer is a decrease in 25% rather than a decrease in 33%. Can anyone help me with this?

 

[unique135]

nvm...

 

[MDub1]

I wonder if this is a case where the book is wrong...

 

[unique135]

I wonder if this is a case where the book is wrong...

This is what I am thinking too. Perhaps, wait for BerkReviewTeach to reply on this.

 

[Vanguard23]

Yeah, I did a simply conservation equation:

p1v1 = p2v2

(1atm)(400ml) = (x)(1200ml)

= (1atm)(400ml)/(1200ml) = x

(1atm)(1/3) = x

Incorrect?

 

[BerkReviewTeach]

TBR Chemistry Lecture 6

#40. For lungs with a tidal volume of 400 mL and a total volume following normal expiration of 1200 mL, how does internal pressure change to cause inspiration?

A. It increases by 33%
B. It increases by 25%
C. It decreases by 25%
D. It decreases by 33%


I have been having some trouble with percentage problems. I cannot understand how the answer is a decrease in 25% rather than a decrease in 33%. Can anyone help me with this?

The volume increases from 1200 mL to 1600 mL for a ratio of 4:3 for V2 to V1.

The question only refers to pressure, so we can assume that with PV = k, that the ratio of P2 to P1 must be 3:4. This means that the pressure drops by 1/4, which is 25% of the original pressure.

These ratio/percentage questions can be tricky because of the wording.

 

[Vanguard23]

The volume increases from 1200 mL to 1600 mL for a ratio of 4:3 for V2 to V1.

The question only refers to pressure, so we can assume that with PV = k, that the ratio of P2 to P1 must be 3:4. This means that the pressure drops by 1/4, which is 25% of the original pressure.

These ratio/percentage questions can be tricky because of the wording.

So the OP just made a typo(an INCREASE of 400mL) and the volumes are really 1200mL -> 1600mL?
Because that's logical: a 25%^ in V leads to a 25% drop in P.

 

[MDub1]

The volume increases from 1200 mL to 1600 mL for a ratio of 4:3 for V2 to V1.

The question only refers to pressure, so we can assume that with PV = k, that the ratio of P2 to P1 must be 3:4. This means that the pressure drops by 1/4, which is 25% of the original pressure.

These ratio/percentage questions can be tricky because of the wording.

Oh, ok, you're right. The pressure would drop by 25% in this case when the volume went up by 33%. That is tricky.

 

[Vanguard23]

Oh, ok, you're right. The pressure would drop by 25% in this case when the volume went up by 33%. That is tricky.

Nah, just:

p1v1 = p2v2

p1(1200ml) = p2(1600ml)

p1(1200ml)/(1600ml) = p2 -> p1(3/4) = p2
Pressure2 being 3/4 of p1; dropped by 25%.

 

[unique135]

So the OP just made a typo(an INCREASE of 400mL) and the volumes are really 1200mL -> 1600mL?
Because that's logical: a 25%^ in V leads to a 25% drop in P.

OP didn't made a typo

Tidal volume = 400 mL
Total volume during Normal Expiration = 1200 mL........(basically where tidal volume = 0)
Total volume during Normal Inspiration = 1200 mL + 400 mL = 1600 mL


Your volume is not increasing by 25% but 33%.

Oh, ok, you're right. The pressure would drop by 25% in this case when the volume went up by 33%. That is tricky.

True, this is what got me - thinking straight in percent.
Volume increases by 33%, so pressure has to increase by 33%.

Is it safe to think in ratios?

Volume increases to 4/3, so pressure decrease to 3/4.

 

[Vanguard23]

OK, I see now.

 

[BerkReviewTeach]

True, this is what got me - thinking straight in percent.
Volume increases by 33%, so pressure has to increase by 33%.

Is it safe to think in ratios?

Volume increases to 4/3, so pressure decrease to 3/4.

I said something in my previous post, but I'll write this a second time with the confession that if I wrote it the way I wanted to, there would be about four or five profanities in the sentence.

"These (fit two in here) ratio/percentage questions can be (fit one in here) tricky because of the (fit one or two in here) wording." Yeah, that seems more accurate.

 

[SirWilliamOsler]

Thanks for all the replies! I think I'm going to stick with the ratio approach. I would like to add another question that I wanted some clarification on.

TBR Lecture 7 Passage 6 question #43

#43. Aliquot 1 from the mixture has which of the following mole distribution percentages?

A. 50.0% acetophenone and 50.0% 2-propanol
B. 33.3% acetophenone and 66.7% 2-propanol
C. 20.0% acetophenone and 80.0% 2-propanol
D. 11.1% acetophenone and 88.9% 2-propanol

The answer is choice D and I chose answer choice B. Are the percentages in answer choice D referring to the mole fractions in vapor while the percentages in answer choice B referring to the mole fractions in the solution? The explanation in the book did not give a reason as to why answer choice B was wrong. This is another one of those percentage problems that I am having trouble with!!

 

[SirWilliamOsler]

retracted...

 

[Oh_Gee]

The volume increases from 1200 mL to 1600 mL for a ratio of 4:3 for V2 to V1.

The question only refers to pressure, so we can assume that with PV = k, that the ratio of P2 to P1 must be 3:4. This means that the pressure drops by 1/4, which is 25% of the original pressure.

These ratio/percentage questions can be tricky because of the wording.

is the ratio P2 : P1 3:4 because thats the inverse of V2:v1? I'm still confused about the pressure drop. can you please explain in terms of algebra how you got 25%?