How To Find Work If I have Pressure and Volume

In summary, to calculate the work done in an adiabatic process, you would use the equation P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}, where \gamma is the adiabatic index (in this case, 1.4). This process does not involve heat exchange, so the formula for isothermal processes would not be applicable. To calculate the work done in a cycle, you would use the formula Workdone/cycle = n/n-1 x P1V1 x [ (P2/P1)power n-1/n - 1 ], and convert the final result to Watts by dividing by 60.
  • #1
EngNoob
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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.
 
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  • #2
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...
 
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  • #3
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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