Ideal Gas Law - Differential Approximation due to variable increase in %.

Click For Summary
The discussion centers on using the ideal gas law, P=kT/V, to approximate the percentage change in pressure due to a 3% increase in temperature and a 5% increase in volume. The original poster calculated a 2% change in pressure, while an online solution indicated a 4% change, leading to confusion about the discrepancy. The poster suspects a misunderstanding related to a division by 1/2 in the online solution. Clarification is sought on the logic behind this division and the correct application of differentials in the problem. Understanding the differential approximation is crucial for accurately determining the percentage change in pressure.
Beamsbox
Messages
61
Reaction score
0

Homework Statement


This is a problem form my calculus book, which states:

According to the ideal gas law, the pressure, temperature, and volume of a confined gas are related by P=kT/V, where K is a constant. Use differentials to approximate the percentage change in pressure if the temperature of a gas is increased 3% and the volume is increased 5%.

The Attempt at a Solution



Here, I include my original attempt at the solution and a solution that I found online. In my original solution, I found the answer to be 2%, but the solution says 4%. I'm not sure why... but I think it's related to something I don't understand in the actual solution, where it seems to me, they divide by 1/2 for no apparent logical reason.

Could someone please explain to me where my logic is flawed?
attachment.php?attachmentid=40470&d=1319929174.jpg


Here is a solution I found online to the same problem, perhaps it's more clear.
attachment.php?attachmentid=40468&d=1319927946.jpg
 

Attachments

  • calc_diff.jpg
    calc_diff.jpg
    19.5 KB · Views: 2,494
  • Calc_Diff_hand.jpg
    Calc_Diff_hand.jpg
    52.2 KB · Views: 1,246
Last edited:
Physics news on Phys.org
Edited to make images easier to see. (Sorry.)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
10K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
Replies
1
Views
2K
  • · Replies 109 ·
4
Replies
109
Views
8K