How To Find Work If I have Pressure and Volume

Click For Summary
To calculate work done in the given scenario, the isothermal process formula is incorrect due to the provided expansion/compression index of 1.4, indicating an adiabatic process for a diatomic gas. The correct relationship for an adiabatic process is P1V1^γ = P2V2^γ, where γ is the adiabatic index. Work can be calculated using the formula for work done in a cycle, which involves integrating pressure with respect to volume. It is important to convert pressure to kN/m² for accurate results in kJ/minute, which can then be converted to power in kW. Understanding the distinction between isothermal and adiabatic processes is crucial for accurate calculations.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 23 ·
Replies
23
Views
1K
Replies
12
Views
1K
  • · 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
857
  • · Replies 2 ·
Replies
2
Views
3K