## 4A.3

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

kendal mccarthy
Posts: 109
Joined: Wed Nov 14, 2018 12:22 am

### 4A.3

Can someone explain how they calculated w for part a given "the inner diameter of the pump is 3.0 cm and the pump is depressed to 20.0cm with a pressure of 2.00atm" ?

RasikaObla_4I
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

### Re: 4A.3

Because work is equal to -p delta V, to find the volume of the pump, you need to use the formula for calculating the volume of a cylinder which is pi (r)^2 X height, with the height being -20 because the pump is being compressed. Using the volume and the given pressure, you can plug the values into the equation to find the work.

Posts: 104
Joined: Fri Aug 09, 2019 12:16 am

### Re: 4A.3

Because the problem describes an irreversible expansion with a constant external pressure, you use the equation w=-P(ex)(deltaV). The delta V is equal to the area of the region compressed or expanded times the distance compressed. Therefore the diameter given is used to find the area of the region compressed and you plug in the rest of the given numbers.

Ryan 1K
Posts: 101
Joined: Fri Aug 09, 2019 12:15 am

### Re: 4A.3

Adding on to the previous replies, the equation $w=-P\Delta V$ outputs an answer with units m^3 $\cdot$ atm. To convert this answer to J, you need to convert to m^3 $\cdot$ Pa, which would involve multiplying your original answer by 101.325.

Ruby Richter 2L
Posts: 103
Joined: Thu Jul 25, 2019 12:17 am

### Re: 4A.3

What would be the change in volume for this problem then if final is what you get from finding how much the pump was compressed?