Discussion
In a graph with Helmholtz function on the \(y\)-axis and volume on the \(x\)-axis, the slope of the isothermal curves for a finite volume system containing an ideal gas is
Answer: D
The slope is finite, negative, and non-zero.
From \(dA = -S\,dT - P\,dV\), the slope of an isotherm on an A-V diagram is
\[ \left(\frac{\partial A}{\partial V}\right)_T = -P \]
For an ideal gas at finite volume, \(P = nRT/V\) is positive and finite, so the slope is negative, finite and non-zero, and it flattens as V grows.
A zero slope would need zero pressure, an infinite one would need infinite pressure, and a positive slope would need negative pressure, none of which applies to an ideal gas.
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