## 4A.3

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

### 4A.3

Air in a bicycle pump is compressed by pushing in the handle. The inner diameter of the pump is 3.0 cm and the pump is depressed 20. cm with a pressure of 2.00 atm. (a) How much work is done in the compression? (b) Is the work positive or negative with respect to the air in the pump? (c) What is the change in internal energy of the system?

I'm not understanding how to calculate part a of the problem. I used w = P * A * D , but keep getting an answer far off from the correct one.

Jeremy_Guiman2E
Posts: 82
Joined: Fri Sep 28, 2018 12:29 am

### Re: 4A.3

You do use w = P * A * D!

w = 2 atm * (0.015 m)2 * 0.2 m * pi
w = 2.8274 * 10-4 m3 * atm

Then, you can multiply the answer by 101,325 Pa to convert from atm to Pa. Once you have your answer in m3 * Pa, it is equal to that amount in joules (J).

kausalya_1k
Posts: 50
Joined: Wed Nov 14, 2018 12:23 am

### Re: 4A.3

Should we always convert to J (using the 101325 Pa value) on exams? Correct me if I'm wrong, but I don't recall hearing that in lecture?

rachelle1K
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

### Re: 4A.3

I did this problem yet I keep getting my final answer as 28.6, which would round up to 29. The answer says it' supposed to be 28. Has anyone else been getting the same thing?

rachelle1K
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

### Re: 4A.3

kausalya_1k wrote:Should we always convert to J (using the 101325 Pa value) on exams? Correct me if I'm wrong, but I don't recall hearing that in lecture?

I didn't hear him talk about it either in lecture, but it makes sense because work is equal to J and you have to make sure the units work out. Pa is equal to J.m^-3, and by multiplying by m^2 (area) and m (distance), we are left with the joules.

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

### Re: 4A.3

Jeremy_Guiman2E wrote:You do use w = P * A * D!

w = 2 atm * (0.015 m)2 * 0.2 m * pi
w = 2.8274 * 10-4 m3 * atm

Then, you can multiply the answer by 101,325 Pa to convert from atm to Pa. Once you have your answer in m3 * Pa, it is equal to that amount in joules (J).

Where did you get 0.015m from and pi from?

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### Re: 4A.3

DesireBrown1J wrote:
Jeremy_Guiman2E wrote:You do use w = P * A * D!

w = 2 atm * (0.015 m)2 * 0.2 m * pi
w = 2.8274 * 10-4 m3 * atm

Then, you can multiply the answer by 101,325 Pa to convert from atm to Pa. Once you have your answer in m3 * Pa, it is equal to that amount in joules (J).

Where did you get 0.015m from and pi from?

I believe it's to convert the diameter (3.00cm) to area!

Diameter = 3.00 cm
Radius = 3.00 cm/ 2 = 1.5 cm = 0.015 m (Joules is m2.kg.s-2, so you need to convert cm to m)

Area = pi*0.015m2

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### Re: 4A.3

rachelle1K wrote:I did this problem yet I keep getting my final answer as 28.6, which would round up to 29. The answer says it' supposed to be 28. Has anyone else been getting the same thing?

I got that too, but when I used 2.8*10-4 instead of 2.8274*10-4, I got 28.371, which rounds down to 28. Are we supposed to do it like this? I need to review significant figures.

Zoe Gleason 4F
Posts: 51
Joined: Tue Jul 23, 2019 12:15 am

### Re: 4A.3

rachelle1K wrote:I did this problem yet I keep getting my final answer as 28.6, which would round up to 29. The answer says it' supposed to be 28. Has anyone else been getting the same thing?

I also got 29. 28 is the product of using sig figs early, then rounding down. I think either answer would be correct, but my TA did tell us to not round until the very end (which would produce 29).