Calculate the entropy change when a. two moles of H2O(g) are cooled irreversible at constant p from 120°C to 100°C. b. one mole of H2O(g) is expanded at constant pressure of 2 bar from an original volume of 20 L to a final volume of 25 L. You can consider the gas to be ideal. c. one hundred grams of H2O(s) at -10°C and 1 bar are heated to H2O(l) at +10°C and 1 bar.

The answers to the question are

a. S₂ - S₁  = 7.95 J/K

b. S₂ - S₁  = -16.9 J/K

c. S₂ - S₁  = -30.78 J/k

Explanation:

a. To solve the question, we note that

The change in entropy S₂ - S₁ is given by

Where

T₁ = Initial temperature in Kelvin = 120 + 273.15 K = 393.15 K

T₂ = Final temperature in Kelvin = 100 + 273.15 K = 373.15 K

and Cp for water = 4.23 kJ/(kg·K)

Molar mass of water = 18 g/mol

Therefore mass of water = 2*18 =36 grams = 0.036 kg

The change in entropy then is = 0.00795 kJ/(K)

= 7.95 J/K

b.  At constant pressure for one  mole = 18 grams

Therefore change in entropy S₂ - S₁ =

= = -0.01686965 kJ/(K) = -16.9 J/K

c.  100 grams of H₂O from -10 °C (263.15 K) to 10 °C (283.15)

100 g = 0.1 kg

we have S₂ - S₁ is given by = = -0.03078 kJ/K

= -30.78 J/k

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

all eukaryotic cell have mitochondria : )