Answer :
The pressure in a 9.00 L tank with 36.2 grams of fluorine gas at 355 K is 11.9 atm.
The Ideal Gas Law can be used to find the pressure of fluorine gas in the tank at 355 K.
PV = nRT,
where P is pressure,
V is volume,
n is the number of moles of gas,
R is the gas constant, and T is temperature in Kelvin.
The ideal gas law formula for determining the pressure of a gas in a container can be rearranged to solve for pressure.P = nRT / V
The number of moles of fluorine gas can be found using the formula:n = m / MM
where n is the number of moles,
m is the mass of the gas, and MM is the molar mass of the gas.
In this case, the mass of the gas is 36.2 grams, and the molar mass of fluorine is 18.998 g/mol, so:
n = 36.2 g / 18.998 g/mol = 1.91 mol
Substituting values in the formula:P = nRT / V = (1.91 mol)(0.0821 L*atm/mol*K)(355 K) / 9.00 L= 11.9 atm
Therefore, the pressure in a 9.00 L tank with 36.2 grams of fluorine gas at 355 K is 11.9 atm.
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Final answer:
To find the pressure of fluorine gas in a tank, convert the mass of gas to moles and use the ideal gas law PV=nRT. For 36.2 grams of fluorine gas at 355 K in a 9.00 L tank, the pressure is calculated as 3.115 atm.
Explanation:
The question is asking for the pressure of fluorine gas in a 9.00 L tank at 355 K. To solve this, we can use the ideal gas law equation PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. First, we need to convert the mass of fluorine gas to moles by using the molar mass of fluorine (F2), which is approximately 38 g/mol (19 g/mol for each F atom).
Number of moles (n) = 36.2 g / 38 g/mol = 0.953 moles.
Using the ideal gas constant R = 0.0821 L·atm/(K·mol), and the given temperature and volume, we can rearrange the equation to solve for pressure (P):
P = nRT/V
P = (0.953 mol)(0.0821 L·atm/(K·mol))(355 K) / (9.00 L)
P = 3.115 atm
