4A.3 7th edition


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

4A.3 7th edition

Postby Maggie Doan 1I » Wed Jan 30, 2019 6:54 pm

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

Postby Elizabeth Gallmeister 1A » Wed Jan 30, 2019 7:43 pm

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

Postby Elizabeth Gallmeister 1A » Wed Jan 30, 2019 7:45 pm

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.


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