If the vapor pressure of ethanol at 34.7degree C is 100
1 answer:
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
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Explanation:
Apply the mass of balance as follows.
Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank
[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] =
+ h = 1
= 1 - h
= dt
= t + C
Given at t = 0 and V = 0
= 0
or, h = 0
-ln(1 - h) = t + C
Initial condition is -ln(1) = 0 + C
C = 0
So, -ln(1 - h) = t
or, t = ........... (1)
(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.
t =
t =
=
= 0.916 seconds
(b) As maximum height of water level in the tank is achieved at steady state that is, t = .
1 - h = exp (-t)
1 - h = 0
h = 1
Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.