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Alex73 [517]
3 years ago
6

Express the volume expansivity and the isothermal compressibility as functions of density rho and its partial derivatives. For w

ater at 50°C and 1 bar, κ = 44.18 × 10 −6 bar −1 . To what pressure must water be compressed at 50°C to change its density by 1%? Assume that κ is independent of P.
Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

Explanation:

κ = 1/v  x dv/dp

= 1/ρ x (+δρ/ δp )_T

δρ = 1 , ρ = 100

44.18 x 10⁻⁶ = ( 1/100) x( 1 / δp)

δp = 10⁻² / (44.18 x 10⁻⁶ )

= .0226 x 10⁴

= 226 bar

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2 years ago
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-
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\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\

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5 0
3 years ago
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ratelena [41]

Answer:

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3 years ago
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