Ideal monatomic gas engine problem

  • Thread starter Thread starter ken12380
  • Start date Start date
  • Tags Tags
    Engine Gas
Click For Summary
SUMMARY

The discussion focuses on calculating the total heat added during a thermodynamic cycle of an ideal monatomic gas in a cylindrical container. The key parameters include pressures PA = PB = 12.0 atm, PC = PD = 6.0 atm, volumes VA = VD = 5.0 m3, VB = VC = 30.0 m3, and a temperature of point C at 900 K with 2437.3376 moles of gas. The correct approach involves using the equations Q=nCpΔT and Q=nCvΔT, focusing solely on the heat added to the gas during pressure increases at constant volume and expansions at constant pressure.

PREREQUISITES
  • Understanding of thermodynamic cycles
  • Familiarity with the ideal gas law (PV=nRT)
  • Knowledge of specific heat capacities (Cp and Cv)
  • Ability to perform calculations involving moles and temperature changes
NEXT STEPS
  • Study the derivation and application of the ideal gas law (PV=nRT)
  • Learn about the differences between isochoric and isobaric processes
  • Explore the concept of heat transfer in thermodynamic systems
  • Investigate the calculations of work done in thermodynamic cycles
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, mechanical engineers, and anyone involved in the analysis of heat engines and gas behavior in confined systems.

ken12380
Messages
1
Reaction score
0

Homework Statement


An ideal monatomic gas is confined to a cylindrical container having a movable, yet "snuggly-fitting," lid that slides up and down to accommodate changes in pressure and volume. The diagram shown below, outlines the steps in a heat cycle starting at position A and moving through positions B, C, D and then back to A.
2a7246d7-f2d0-4e94-bdb2-45bb48bd08a9.gif

*ignore the axes labels*
PV diagram2
Given:
PA = PB = 12.0 atm
PC = PD = 6.0 atm
VA = VD = 5.0 m3
VB = VC = 30.0 m3
temperature of point C is 900 K
moles of gas are present 2437.3376

How much total heat is added during the cycle?

Homework Equations


Q=nCp delta T
Q=nCv delta T
PV=nRT

The Attempt at a Solution


I tried:
Q=Q+Q=nCp delta T+nCv delta T=2437.3376*20.78*150+2437.3376*12.47*1500
=53187581 J but this is wrong.
Why? and what is the actual answer?
 

Attachments

  • 2a7246d7-f2d0-4e94-bdb2-45bb48bd08a9.gif
    2a7246d7-f2d0-4e94-bdb2-45bb48bd08a9.gif
    3.7 KB · Views: 459
Physics news on Phys.org


You have to determine where heat is being added. It is being added(flowing into the gas) when the gas pressure increases at constant volume and when the gas expands at constant pressure. Heat flows out when it loses pressure at constant volume and when it is being compressed at constant pressure. You have to calculate only the heat flow INTO the gas.

AM
 

Similar threads

Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
1K
Replies
3
Views
5K
Replies
2
Views
12K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K