## 4A 3

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

ckilkeary 2G
Posts: 68
Joined: Fri Aug 09, 2019 12:16 am

### 4A 3

This question states that 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?

First of all, I do not know where to start and which equation to use. At first I thought it was w= - Pex * ΔV but I can't see how this would be used to find the work properly or if I'm using the right equation.

Jielena_Bragasin2G
Posts: 104
Joined: Sat Aug 24, 2019 12:18 am
Been upvoted: 1 time

### Re: 4A 3

That is indeed the correct equation that you are thinking of using! You need to use the dimensions (given in centimeters) to find the volume of the compression. with that you can find if the work is positive or negative and the change in internal energy.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: 4A 3

After you calculate the dimension change of the pump, you multiply by the external pressure, which is given as 2 atm. Remember to convert this to pascals, whether that is before or after the calculation. This should give you an answer of 28.

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

### Re: 4A 3

BeylemZ-1B wrote:After you calculate the dimension change of the pump, you multiply by the external pressure, which is given as 2 atm. Remember to convert this to pascals, whether that is before or after the calculation. This should give you an answer of 28.

Why must you convert the answer to pascals when the units (J) does not include pascals?

Max Madrzyk Dis 4G
Posts: 102
Joined: Wed Sep 18, 2019 12:21 am

### Re: 4A 3

You have the right equation but you need to figure out what delta V is. To do that use delta V= -pi*r^2*d, then plug it into the equation you have and use the pressure given.

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

### Re: 4A 3

Why must you convert the answer to pascals when the units (J) does not include pascals?

You must convert the answer to Pascals because the textbook gives a conversion factor that converts meters cubed x Pascals into joules (1 Pa x m3 = 1 J). Remember that 1 atm = 101,325 Pa.