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bonufazy [111]
3 years ago
8

An object of size 5cm is placed in front of a concave mirror of focal length of magnitude 20 cm at a distance of 40cm in front o

f the mirror. The image formed is
Physics
1 answer:
zheka24 [161]3 years ago
7 0

Are there any options?

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Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from
Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

  • Density of the ice, ρ = 917 kg/m³
  • Initial pressure, P₁ = 1.0 bar
  • Final pressure, P₂ = 2.0 bar
  • Temperature, T = - 10 C
  • Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

\Delta G = V(P_2-P_1) \\\\

where;

V is the volume of the ice

Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L

Change in pressure;

P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm

The change in the Gibb's free energy per mole (G);

\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G =  0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

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