## 4A.3 7th edition

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

Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### 4A.3 7th edition

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.
How do you calculate the change in internal energy?

Elizabeth Gallmeister 1A
Posts: 41
Joined: Fri Sep 28, 2018 12:21 am

### Re: 4A.3 7th edition

You're going to want to use the formula we learned in class today: w = -Pex (deltaV).

Plugging in numbers, we get w = -(-2.00 atm) (1.5 cm)^2 (pi) (20 cm)
From that, you get w = 283. atm. cm^3, which you then convert to J by multiplying by (1mL/1cm^3)(1L/1000mL)(101.325J/L.atm) = 29J

Since it's compression, work is going to be positive, and no heat is released, so deltaU = w = 29J

Elizabeth Gallmeister 1A
Posts: 41
Joined: Fri Sep 28, 2018 12:21 am

### Re: 4A.3 7th edition

OH also I think the 7th edition solutions manual is wrong; it says 8 or something?? I'm pretty sure the 6th edition manual is right; it says 29. or maybe 28J.