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How To Find Work If I have Pressure and Volume

  • Thread starter EngNoob
  • Start date
  • #1
38
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.
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
1,648
4
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 [tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex], where gamma is the adiabatic index (in your case, [tex]\gamma = 1.4[/tex]). 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:
  • #3
6
0
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.
 

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