Ideal gas temperature/pressure/volume problem

  • Thread starter Thread starter colbyrandell
  • Start date Start date
  • Tags Tags
    Gas Ideal gas
Click For Summary
SUMMARY

The discussion focuses on solving an ideal gas problem involving a gas expanding in a perfectly insulated cylinder. The initial conditions include a pressure of 38 bar and an internal energy of 1400 kJ, with the gas expanding until its internal energy reaches 1300 kJ. The relevant equations include the expansion law PV1.8 = C and the work done equation W = (PiVi - PfVf)/(n-1). The correct approach to calculate work involves integrating the pressure with respect to volume, leading to W = ∫ViVfP dV.

PREREQUISITES
  • Understanding of the ideal gas law and thermodynamic principles
  • Familiarity with the concept of internal energy in thermodynamics
  • Knowledge of calculus for integrating pressure with respect to volume
  • Experience with the specific heat capacities and their relation to work done in gas expansions
NEXT STEPS
  • Study the derivation of the work done by a gas during expansion using calculus
  • Learn about the implications of adiabatic processes in thermodynamics
  • Explore the application of the first law of thermodynamics in various scenarios
  • Investigate the behavior of real gases compared to ideal gases under different conditions
USEFUL FOR

Students and professionals in physics and engineering, particularly those studying thermodynamics and fluid mechanics, will benefit from this discussion.

colbyrandell
Messages
1
Reaction score
0

Homework Statement


A mass of gas at an initial pressure of 38 bar, and with an internal energy of 1400kJ, is contained in a perfecty insulaed cylinder of volume 0.06 meters cubed. the gas is allowed to expand behind a piston until its internal energy is 1300kJ. the law of expansions is PV^1.8=C. find:
work done
the final volume
the final pressure

Homework Equations


PV^1.8=c
Uinital + Q = U final + W
W = (PiVi - PfVf)/(n-1)
PiVi^1.8=PfVf^1.8

The Attempt at a Solution


Ui - Uf = W - Q
1400-1300= W - 0
1000 = W[/B]
 
Physics news on Phys.org
Welcome to PF!
colbyrandell said:

Homework Statement


A mass of gas at an initial pressure of 38 bar, and with an internal energy of 1400kJ, is contained in a perfecty insulaed cylinder of volume 0.06 meters cubed. the gas is allowed to expand behind a piston until its internal energy is 1300kJ. the law of expansions is PV^1.8=C. find:
work done
the final volume
the final pressure

Homework Equations


PV^1.8=c
Uinital + Q = U final + W
W = (PiVi - PfVf)/(n-1)
PiVi^1.8=PfVf^1.8

Your equation for the work is not correct. The work of the gas when it expands from volume Vi to volume Vf is equal to the integral W=\int _{V_i}^{V_f}{PdV}

In case of this gas, P = \frac{P_iV_i^{1.8}}{V^{1.8} }
colbyrandell said:

The Attempt at a Solution


Ui - Uf = W - Q
1400-1300= W - 0
1000 = W[/B]
1400-1300=100 kJ

ehild
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
10
Views
2K
Replies
16
Views
4K