Monatomic gas, Isochoric Process.

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving a monatomic ideal gas undergoing an isochoric process, where the gas is heated from an initial temperature to a final temperature. Participants are tasked with finding the work done, change in internal energy, and total heat involved in the process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the ideal gas equations and concepts of kinetic energy to determine the change in internal energy and heat transfer. Some participants question the application of the average kinetic energy formula to the entire system rather than a single molecule.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's calculations and clarifying the distinction between the energy of a single molecule and the total energy for the given moles of gas. There is no explicit consensus yet, but guidance has been offered regarding the correct interpretation of the calculations.

Contextual Notes

Participants are navigating the implications of using the average kinetic energy formula in the context of multiple moles of gas, which may lead to confusion regarding the calculations presented by the original poster.

yaylee
Messages
22
Reaction score
0

Homework Statement



n = 1.46 moles of ideal gas are heated isochorically (at constant volume) from tepmerature To = 649 oC to temperature Tf = 1184 oC. Find:

a) Work done on the system.
b) Change in Internal Energy of the system.
c) The total heat, Q, added or removed from the system.

Homework Equations


Change in I.E. = Q + W
KE (avg) = T = 3/2kT, where k = 1.38 x 10^-26 kJ/K

The Attempt at a Solution



a) Since this is an isochoric system, work done is ZERO! (No issues here.)
b,c) Change in Internal Energy, therefore is equal to Q, by the first equation, Change in IE = Q + W.
So, we can calculate change in IE, by change in KE.

KE = (3/2kT) for a monatomic gas.
so, change in KE = 3/2(k)(Tfinal - Tinitial), = (3/2)(1.38E-26)(1500 K - 704K) = change in IE = Q = 1.65 x 10^-23.

Am I reading the problem incorrectly? Many thanks in advance once again!
 
Physics news on Phys.org
(3/2)kT is the average translational KE of just one molecule.
 
It appears you have calculated the change in KE for a single molecule of the gas.
Remember, you have 1.46 moles of gas to start with.
 
Thanks!
 

Similar threads

Internal Energy of an Ideal Gas
  • · Replies 4 ·
Replies
4
Views
3K
Isobaric/isochoric (?) heating of an ideal gas
  • · Replies 5 ·
Replies
5
Views
4K
Heat and work when temperature increases by 1 degree
  • · Replies 21 ·
Replies
21
Views
3K
What Happens When Gas Expands at Constant Pressure?
  • · Replies 12 ·
Replies
12
Views
7K
Isobaric Process: Finding Q, E, W.
  • · Replies 1 ·
Replies
1
Views
7K
PV Diagram of Ideal Monatomic Gas Processes
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Irreversible Isochoric Process in a Cycle
  • · Replies 8 ·
Replies
8
Views
4K
How Is Internal Energy Calculated for an Ideal Gas Using Temperature Change?
  • · Replies 1 ·
Replies
1
Views
2K
Energy of a Gas in equilibrium with BB-radiation
  • · Replies 9 ·
Replies
9
Views
2K
Isobaric Process, finding Change in Internal Energy
  • · Replies 2 ·
Replies
2
Views
3K