# Expansion of an Ideal Gas

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: $$\sum$$Ein - $$\sum$$Eout = (Win - Wout) + (Qin - Qout) = $$\Delta$$U; W = $$\int$$ Fds

Where 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):

Wb = $$\int$$ F*ds

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

Wb = $$\int$$ 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:

Wb = k*$$\int$$ dV/V = k*ln(V2/V1

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 V2 is greater than V1.

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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 = $$\int$$ F*ds

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

W =$$\int$$ p*dV

Here, pressure is constant, so

W = p*$$\int$$ dV = p*$$\Delta$$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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I have a feeling that we're in the same class, seriously.

Anyways...

From our formula sheet:

Work for a compressible substance undergoing isothermal quasi-static process:

W=mRT*ln(V2/V1)

Work for compressible substance undergoing isobaric quasi-static process:

W=p$$\Delta$$V

From there pick some values for p, V1, V2 and T. Use the same values for both equations and see which one results in more work being done.

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