Solving Cylinder Problem: Calculate Work Done in KJ

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The discussion revolves around calculating the work done during the expansion of gas in a piston-cylinder device, where the relationship PV=C applies. The user is confused about how to incorporate the constant n and determine the work done, given the initial and final volumes. Suggestions include using the ideal gas law and the integral for work done, specifically for isothermal processes, as the problem does not specify the type of process. The correct formula for work in this context is W = nRT ln(V_f/V_i), assuming isothermal conditions. The user plans to proceed with this approach despite the lack of clarity on the process type.
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Hello I am having trouble with this problem I was wondering if anyone could help me out here goes,

During some actual expansion and compression process in piston-cylinder devices, the gases are observed to satisfy the relationship PV=C, where n and C are constants. Calculate the work done in KJ when gas expands from 150kPa and 0.03m to a final volume of 0.2m for the case of n = 1.3.

Ok I understand that P1 = 150kpa and V1 = 0.03m and V2 = 0.2m when I try to solve for P2 using P1V1=P2V2 I get 22.5kPa but it doesn't make sense because its expanding and I'm not sure how to incorporate n into the problem, also I am confused as to finding the work done? can anyone hint me in the right direction? thanks.
 
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Does the problem indicate if it is an adiabatic process?

Did you mean to writne PV^n= C instead?

CS
 
no it does not state that it is an adiabatic, I considered that formula but I'm not sure as to how I would use it because there is a beginning volume and an ending volume. I'm still pretty confused it has been a while since Thermo I this is just a review for Thermo II. I appreciate the help thanks!
 
You really need to know the process by which the final state of the gas is obtained, however since you don't know that, let's assume it is isothermal.

The general equation for work done by an ideal gas is...

W = \int_{V_i}^{V_f} PdV

The equation of state for an ideal gas is...

PV = nRT, which once rearranged gives P = \frac{nRT}{V}.

So the work becomes...

W = \int_{V_i}^{V_f} \frac{nRT}{V}dV

Since this is isothermal nRT is constant, which then gives...

W = nRT \int_{V_i}^{V_f} \frac{dV}{V}

After integrating and using a log identity you have...

W = nRT \cdot \ln \frac{V_f}{V_i}

If it is not isothermal then of course the work will be different. Let me know if it is not isothermal.

Hope that helps...

CS
 
it doesn't state that is isothermal ho it is written above is exactly how the problem is written, but I will attempt it that way thank you.
 

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