4A.3 7th Ed


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4A.3 7th Ed

Postby Noh_Jasmine_1J » Tue Jan 29, 2019 9:50 am

how do I figure out how to determine the change in volume when I am solving for work done? the question asks to solve for the work of a pump in compression with a diameter of 3cm and the pump is depressed 20cm with a pressure of 2atm. thank you~

Chloe Thorpe 1J
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Re: 4A.3 7th Ed

Postby Chloe Thorpe 1J » Tue Jan 29, 2019 11:00 am

The change in volume in this question is given by -π*d*r^2. So, it would be -π*(20)*(1.5)^2 = -0.14 cm^3. Then, you multiply that by 1L/1000cm^3 to get it in the right unit, so you have -0.14 L.

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Re: 4A.3 7th Ed

Postby 204929947 » Tue Jan 29, 2019 5:29 pm

why do we need to put -(pi)(d) (r^2)???? where does this formula come from????

Dimitri Speron 1C
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Re: 4A.3 7th Ed

Postby Dimitri Speron 1C » Wed Jan 30, 2019 12:20 pm

Change in volume is (pi)(r^2)(change in d) because r is a constant and does not change in between the final and initial volumes. (at least in this example, it could potentially change but that would be a rather more complicated) So in order to calculate change in volume you just need to replace the d in the equation for volume with a change in d. With this you can calculate the work using the equations from class.

Ryan Danis 1J
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Re: 4A.3 7th Ed

Postby Ryan Danis 1J » Wed Jan 30, 2019 3:57 pm

I was having some trouble with this problem as well and the above replies have helped (I wasn’t squaring r or converting to liters). However I have one question. How come the work here is negative since it appears to be work done on the system instead of the system doing work? Is the system here the handle of the bicycle pump?

Maggie Doan 1I
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Re: 4A.3 7th Ed

Postby Maggie Doan 1I » Wed Jan 30, 2019 6:46 pm

Why is it -πr^2?

Cole Elsner 2J
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Re: 4A.3 7th Ed

Postby Cole Elsner 2J » Wed Jan 30, 2019 7:28 pm

I believe we'll learn how to do this in an upcoming lecture. Don't stress yourself out over it just yet! I'm certain we'll be taught how to do this, if not the book and chemistry community will surely help!

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