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**1. Consider an I.G. expanding from a given state to a fixed final volume at either an isobaric or isothermal process. For which case is the work done greater?**

**2.Energy balance: [tex]\sum[/tex]E**

_{in}- [tex]\sum[/tex]E_{out}= (W_{in}- W_{out}) + (Q_{in}- Q_{out}) = [tex]\Delta[/tex]U; W = [tex]\int[/tex] FdsWhere Q represents heat, W represents work, U represents internal energy, and E represents total energy of the system.

## The Attempt at a Solution

Work done for the expansion of an ideal gas at an

**isothermal**process (all integrals taken from 1 to 2):

W

_{b}= [tex]\int[/tex] F*ds

Where F = p*A, and A*ds = dV;

W

_{b}= [tex]\int[/tex] p*dV

Pressure is decreasing, and using the ideal gas law is a function of volume;

p*V = n*R*T -> p = (n*R*T)/V

Knowing n*R*T to be a constant, and calling this constant k, we have:

W

_{b}= k*[tex]\int[/tex] dV/V = k*ln(V

_{2}/V

_{1}

The example used is a P-C (piston cylinder device), and is a closed system. Here, we note that boundary work will be greater than 1, because V

_{2}is greater than V

_{1}.

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Work done for the expansion of an ideal gas at an

**isobaric**process (again, all integrals taken from 1 to 2):

W = [tex]\int[/tex] F*ds

F = p*A; A*ds = dV, therefore:

W =[tex]\int[/tex] p*dV

Here, pressure is constant, so

W = p*[tex]\int[/tex] dV = p*[tex]\Delta[/tex]V

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I'm not too sure where to go from here, and am also not sure if the above expressions are correct. Picking initial and final conditions to be the same for both functions, however, I find that the work done in an isobaric process is greater than the work done in an isothermal process.

Thanks

DMBdyn

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