having a lesser value by a process of depletion
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
The total volume of water that would be removed will be 75 mL
<h3>Dilution equation</h3>
Using the dilution equation:
M1V1 = M2V2
In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M
Substitute:
V2 = 500 x 10.20/12
= 425 mL
The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:
500 - 425 = 75 mL
More on dilution can be found here: brainly.com/question/7208939
Answer:
Real gas particles have significant volume
Real gas particles have more complex interactions than ideal gas particles.
Explanation:
An ideal gas is an imaginary concept and a gas behaves almost ideally at certain pressure and temperature conditions.
The gas in real deviates from the ideal behavior as some of the assumptions made for ideal gases are not true in case of real gases.
Real gas particles have significant volume as compared to vessel unlike ideal gases.
There are interactions present in between real gas molecules at high pressure conditions.
Periodic table is arranged according to atomic size and other properties.
