Let's assume that the F₂ gas has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
Moles = mass / molar mass
Molar mass of F₂ = 38 g/mol
Mass of F₂ = 76 g
Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol
P = ?
V = 1.5 L = 1.5 x 10⁻³ m³
n = 2 mol
R = 8.314 J mol⁻¹ K⁻¹
T = -37 °C = 236 K
By substitution,
P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K
p = 2616138.67 Pa
p = 25.8 atm = 26 atm
Hence, the pressure of the gas is 26 atm.
Answer is "a".
76 g of fluorine gas in a 1.50 L vessel at -37°C exert a pressure of (a) 26 atm.
We have 76 g of fluorine gas in a closed vessel. First, we will convert 76 g to moles using its molar mass (38.00 g/mol).
[tex]76 g \times \frac{1mol}{38.00g} = 2.0 mol[/tex]
Then, we will convert -37 °C to K using the following expression.
[tex]K = \° C + 273.15 = -37\° C + 273.15 = 236 K[/tex]
Finally, we can calculate the pressure exerted by the gas using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\P = \frac{n \times R \times T}{V} = \frac{2.0mol \times (0.0821atm.L/mol.K) \times 236K}{1.50L} = 26atm[/tex]
76 g of fluorine gas in a 1.50 L vessel at -37°C exert a pressure of (a) 26 atm.
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