Question

What is the pressure in a 9.00 L tank with 92.1 grams of fluorine gas at 355 K?

Answer 1

The pressure in a 9.00 L tank with 92.1 grams of fluorine gas at 355 K is approximately 6.900 atm, calculated using the Ideal Gas Law and the given values for temperature, volume, and mass of the gas.

Step-by-step explanation:

The question is asking to calculate the pressure of a given amount of fluorine gas in a tank at a specified temperature. This falls under the subject of Chemistry and is typically covered at the high school level. We can use the Ideal Gas Law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in Kelvin.

To find the pressure (P), we need to determine the number of moles (n) of fluorine gas first. The molar mass of fluorine (F2) is approximately 38 g/mol, so with 92.1 grams, we have:

n = (92.1 grams) / (38 grams/mol) = 2.423 moles of F2

The volume (V) is given as 9.00 L, and the temperature (T) is 355 K. Using the gas constant R = 0.0821 L·atm/mol·K, we can solve for pressure (P):

P = (nRT) / V

P = (2.423 moles)(0.0821 L·atm/mol·K)(355 K) / (9.00 L)

P = 6.900 atm (rounded to three significant figures)

Therefore, the pressure in the 9.00 L tank with 92.1 grams of fluorine gas at 355 K is approximately 6.900 atm.

Answer 2

To determine the pressure in a tank, we can use the ideal gas law equation, which states that the pressure (P) of a gas is equal to the product of its molar amount (n), the ideal gas constant (R), and its temperature (T), divided by its volume (V). The equation can be written as:

P = (n * R * T) / V

In this case, we are given the volume of the tank (V = 9.00 L), the molar amount of fluorine gas (n = 92.1 g / molar mass of fluorine), and the temperature (T = 355 K). We need to calculate the molar amount of fluorine gas first.

The molar mass of fluorine (F₂) is approximately 38.0 g/mol. Therefore, the molar amount of fluorine gas can be calculated as follows:

n = mass / molar mass

n = 92.1 g / 38.0 g/mol

n ≈ 2.42 mol

Now we can substitute the known values into the ideal gas law equation:

P = (n * R * T) / V

The ideal gas constant (R) is approximately 0.0821 L·atm/(mol·K).

P = (2.42 mol * 0.0821 L·atm/(mol·K) * 355 K) / 9.00 L

Simplifying this expression gives us:

P ≈ 7.61 atm

Therefore, the pressure in the 9.00 L tank with 92.1 grams of fluorine gas at 355 K is approximately 7.61 atm.