The universal gas equation is a representation of the ideal gas law
The amount of water in the vapor phase is approximately 0.133 grams
The reason the above values is correct is given as follows:
The given parameters are;
The mass of liquid water = 1.0 g
Volume of the container the eater is enclosed, V = 1.00 liter
Pressure in the container, P = 134.4 mmHg
Temperature of the vapor, T = 19°C = 292.15 K
Assumption: The vapor is an ideal gas
Solution:
The universal gas equation is presented as follows;
P·V = n·R·T
Therefore;
[tex]n = \dfrac{P \cdot V}{R \cdot T}[/tex]
Where;
n = The number of moles
R = The universal gas constant = 62.363 mmHg·L·mol⁻¹·K⁻¹
Therefore;
[tex]n = \dfrac{134.4 \times 1.00}{62.363 \times 292.15} \approx 0.007376772[/tex]
The molar mass of water, H₂O, MM = 18.01528 g/mol
Mass of the vapor, m = n × MM
∴ The mass of the vapor, m = 0.007376772 × 18.01528 ≈ 0.133
The amount of water (in g) un the vapor phase, m ≈ 0.133 g
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