Question

Water at 1 atm pressure is compressed to 430 atm pressure isothermally. Determine the increase in the density of water. Take the isothermal compressibility of water to be 4.80 × 10−5 atm−1. The density of water at 20°C and 1 atm pressure is rho1 = 998 kg/m3.

Answer 1

Answer:

Explanation:

compressibility = 1 / bulk modulus of elasticity ( B )

B = 1 / 4.8 x 10⁻⁵ = Δp / Δv /v ( Δp is change in pressure , Δv is change in volume )

1 / 4.8 x 10⁻⁵ = 429  / Δv /v

Δv /v = 429 x 4.8 x 10⁻⁵

= 2059.2 x 10⁻⁵

= .021

v = m / d ( d is density and m is mass of the water taken )

taking log and then differentiating

Δv /v  = - Δd / d

- .021 = - Δd / d

Δd =.021  x d

= .021 x 998

= 20.9

new density

= 998 + 20.9

1018.9 kg/m3