Internal Energy: gas inside piston

AI Thread Summary
The discussion centers on calculating the change in internal energy of a gas in a cylinder with a piston under specific conditions. The gas is initially at 300K and heated to 400K, with a heat capacity of 500J/K. The user calculates the heat added (Q) as 50,000 J using the formula Q = C * delta T. They also attempt to determine the internal pressure and the work done (W) by the piston, considering the weight and cross-sectional area. Ultimately, the user realizes they have solved the problem, indicating a successful conclusion to their calculations.
cokeaddict
Messages
2
Reaction score
0
A cylinder (cross section is 0.2m2) with a free moving piston is filled with gas. The piston is attached to a heavy weight W = 10000N. Outside the cylinder, the air is at 300K and 1atm. Initially the gas is at 300K, then it is heated to 400K. The heat capacity of the gas under the constant pressure is 500J/K.

If the length of the gas in the cylinder l increases by 20cm during the heating, find the change in the internal energy of the gas in Joule J.


Im thinking:

C = Q / delta T

C (delta T) = Q

500 x (400-300) = 50000 J = Q

so...delta U = Q - W

W = P delta V

i guess P is the pressure inside the piston...I've told something like this:

Inside pressure = outside pressure - weight_force/cross_sectional_area

I tried that, i came up with 456500 = ( 101300 - 10000) / .2

? i don't know...any help would be appreciated...:frown:
 

Attachments

  • showme.gif
    showme.gif
    3.4 KB · Views: 563
Physics news on Phys.org
oops, i actually got it...thanks though...
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top