How to calculate work done with change in volume and pressure.
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Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = -
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P 1 = 100 mmHg
; T 1 = 34.7 °C = 307.07 K
P 2 = 760mmHg
T 2 =T⁻²=?
Δ vap H = 38.6 kJ/mol
R = 0.008314 kJ⋅K -1 mol -1
ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
The amount of heat required to convert H₂O to steam is : 382.62 kJ
<u>Given data :</u>
Mass of liquid water ( m ) = 150 g
Temperature of liquid water = 43.5°C
Temperature of steam = 130°C
<h3 /><h3>Determine the amount of heat required </h3>The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )
where ;
q1 = heat required to change Temperature of water from 43.5°C to 100°C . q2 = heat required to change liquid water at 100°C to steam at 100°C
q3 = heat required to change temperature of steam at 100°C to 130°C
- For q1
M* S*ΔT
= 150 * 4.18 * ( 100 - 43.5 )
= 35425.5 J
- For q2
moles * ΔHvap
= (150 / 18 )* 40.67 * 1000
= 338916.67 J
- For q3
M * S * ΔT
= 150 * 1.84 * ( 130 -100 )
= 8280 J
Back to equation ( 1 )
Amount of heat required = 35425.5 + 338916.67 + 8280 = 382622.17 J
≈ 382.62 kJ
Hence we can conclude that The amount of heat required to convert H₂O to steam is : 382.62 kJ.
Learn more about Specific heat of water : brainly.com/question/16559442