Hi everyone, i was wondering if anyone could help me solve number 3 from the homework problems.
Question 3: Air in a bicycle pump is compressed by pushing in the handle. If 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?
I have no idea on how to start this problem if anyone can help i appreciate it and thank you in advance.
Chapter 8 Question #3 (The First Law)
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Re: Chapter 8 Question #3 (The First Law)
First, I would write down what you're given in variables, ex. Pressure (P) =2.00atm. Then, relate the given with what you're looking for. In part a, you're looking for "work...done in the compression". Recall that work is force over a distance and pressure is force over an area. Page 22 in the course reader, which talks about expansion, would be helpful.
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Re: Chapter 8 Question #3 (The First Law)
a) I would start with the equation w=-Pex*ΔV, where P is given as 2.00 atm. Then substitute ΔV for -A*d. The radius of the pump is 3.0cm/2=1.5cm and the pump is depressed 20.o cm thus, d=20.0cm. Plug these values into the equation ΔV=-A*d and the answer into liters. Once you get ΔV plug it into the equation for work and covert the final answer into joules.
b) Think about how work is done on the air in compression.
c) Use the equation ΔU=q+w. Remember that no heat is transferred so q=0 and substitute your answer from part a for w to find ΔU.
Hope this helps!
b) Think about how work is done on the air in compression.
c) Use the equation ΔU=q+w. Remember that no heat is transferred so q=0 and substitute your answer from part a for w to find ΔU.
Hope this helps!
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