How To Find Work If I have Pressure and Volume

Click For Summary
SUMMARY

The discussion focuses on calculating work done during an adiabatic process for a diatomic gas, given specific pressures and volumes. The correct formula for work in this context is derived from the relationship P1V1γ = P2V2γ, where γ (gamma) equals 1.4. The user initially attempted to apply the isothermal work formula, W = P2V2ln(V2/V1), which is incorrect for this scenario. The correct approach involves integrating P dV and using the formula for work done in a cycle, which incorporates the pressures and volumes provided.

PREREQUISITES
  • Understanding of thermodynamic processes, specifically adiabatic and isothermal processes.
  • Familiarity with the ideal gas law and the concept of the adiabatic index (γ).
  • Knowledge of calculus for integrating pressure-volume work.
  • Ability to convert units, particularly pressure from Pa to kN/m2 for calculations.
NEXT STEPS
  • Study the derivation of the work done formula in adiabatic processes for diatomic gases.
  • Learn how to apply the ideal gas law in conjunction with the adiabatic process equations.
  • Explore the concept of the adiabatic index and its significance in thermodynamics.
  • Practice unit conversions, especially between pressure units and energy units (kJ/min to kW).
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone involved in energy calculations related to gas processes will benefit from this discussion.

EngNoob
Messages
32
Reaction score
0
I have

P_1 = 1 x 10^5 Pa
p_2 = 11.764x10^5 Pa

V_1 0.826 m^3
V_2 0.142 m^3

and

Expansion / Compression Index of 1.4

I need to calculate Work Done.

First of all i identify 1-2 as an Isothermal Process, which gives equation.

W = P_2 * V_2 * ln (V_2 / V_1)

If i use that formula i get an answer.

What i am unsure of is

1) Am i using the right formula based on what data i have, i only have data that i have above.

2) How and what do i use expansion and compression index for?

Thanks for any help, needed urgent to solve a question.
 
Physics news on Phys.org
If you are given an index of 1.4, you are probably supposed to be looking at an adiabatic process for a diatomic gas. So you don't want to use the work formula for an isothermal process.

In a (reversible) adiabatic process, the relationship is P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}, where gamma is the adiabatic index (in your case, \gamma = 1.4). In such a process, Q = 0 ; you would still calculate work by integrating P dV between the two values for volume, but the result is now a little more complicated...
 
Last edited:
well if u r talking abt the work done in a cycle then it is

Workdone/cycle = n/n-1 x P1V1 x [ (P2/P1)power n-1/n - 1 ]

but put pressure in KN/m2 this will give u the answer in KJ/minute, and if u divide the answer by 60 then it will b Kilo-Joules/Sec that is Watts, so that will b the Power in KW.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 23 ·
Replies
23
Views
1K
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
902
  • · Replies 2 ·
Replies
2
Views
3K