(a) Explanation Given: The ideal gas equation is, P V = k T where, P is pressure, V is volume, T is temperature and k > 0 is constant. Theorem used: “Let F be differentiable on its domain and suppose that F ( x , y ) = 0 defines y as a differentiable function of x . Provided F y ≠ 0 , d y d x = − ∂ F ∂ x ∂ F ∂ y = − F x F y ”. Calculation: The given function is P V = k T . Let P be the function of V and T . Take partial derivative on P V = k T with respect to V . P V = k T ∂ P ∂ V ⋅ ( V ) + P ⋅ ∂ V ∂ V = 0 ∂ P ∂ V ⋅ ( V ) + P ⋅ 1 = 0 ∂ P ∂ V = − P V Thus, the value of ∂ P ∂ V is − P V (b) To determine To show: The equation ∂ P ∂ V ⋅ ∂ T ∂ P ⋅ ∂ V ∂ T = − 1 .

Calculus: Early Transcendentals, 2nd Edition

2nd Edition

ISBN: 9780321965165

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 12.5, Problem 61E

(a)

To determine

To compute: The values ∂P∂V,∂T∂P and ∂V∂T .

(b)

To determine

To show: The equation ∂P∂V⋅∂T∂P ⋅∂V∂T=−1 .

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Chapter 12.5, Problem 60E

Chapter 12.5, Problem 62E

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Chapter 12 Solutions

Calculus: Early Transcendentals, 2nd Edition

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The values ∂ P ∂ V , ∂ T ∂ P and ∂ V ∂ T .

Solution Summary: The author explains that the ideal gas equation is PV=kT.

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To compute: The values ∂P∂V,∂T∂P and ∂V∂T .

To show: The equation ∂P∂V⋅∂T∂P ⋅∂V∂T=−1 .

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Chapter 12 Solutions