Answer:
The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C
Explanation:
Here we make use of the Clausius-Clapeyron equation;
Where:
P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K
P₂ = 0.2 atm = The substance vapor pressure at temperature T₂
= The heat of vaporization = 28.5 kJ/mol
R = The universal gas constant = 8.314 J/K·mol
Plugging in the above values in the Clausius-Clapeyron equation, we have;
T₂ = 440.37 K
To convert to Celsius degree temperature, we subtract 273.15 as follows
T₂ in °C = 440.37 - 273.15 = 167.22 °C
Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.
Answer:
Explanation:
From the given information:
The density of O₂ gas = 
here:
P = pressure of the O₂ gas = 310 bar
= 
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴
= 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:
where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:
After solving;
V = 0.1152 L
∴
= 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.
The limiting reactant is the substance of the reactants that runs out first. The excess reactant is the substance with the greater amount than necessary to react completely with the limiting reactant after a chemical reaction.
