On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line (∂P/∂V)s and the slope of the reversible isothermal line (∂P/∂V)T are related as (where, y = Cp/Cv) ?
Correct answer: C. (∂P/∂V)S = y(∂P/∂V)T
- A. (∂P/∂V)S = (∂P/∂V)T
- B. (∂P/∂V)S = [(∂P/∂V)T]Y
- C. (∂P/∂V)S = y(∂P/∂V)T
- D. (∂P/∂V)S = 1/y(∂P/∂V)T
Explanation
For an ideal gas, the isothermal slope is −P/V, while the reversible adiabatic slope is −γP/V. Thus the adiabatic curve is γ times as steep as the isothermal curve at their intersection.
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