A 2.01 L flask containing 2.31 g of helium gas (molar mass = 4.003 g/mol) is separated by a stopcock from a 3.16 L flask containing 3.08 g of fluorine gas (molar mass = 38 g/mol)

A 2.01 L flask containing 2.31 g of helium gas (molar mass = 4.003 g/mol) is separated by a stopcock from a 3.16 L flask containing 3.08 g of fluorine gas (molar mass = 38 g/mol).
After the stopcock is opened the total pressure is 625 mm Hg.
What was the pressure (reported to three significant figures) of the helium before the stopcock was opened?

The correct answer and explanation is :

To solve for the pressure of helium before the stopcock is opened, we can use the ideal gas law and the relationship between pressure, volume, and moles of gas.

Step 1: Calculate the moles of each gas.

We are given:

Mass of helium = 2.31 g
Molar mass of helium = 4.003 g/mol
Mass of fluorine = 3.08 g
Molar mass of fluorine = 38.0 g/mol

To find the moles of each gas, we use the formula:

[
\text{moles} = \frac{\text{mass}}{\text{molar mass}}
]

Moles of helium:

[
n_{\text{He}} = \frac{2.31 \, \text{g}}{4.003 \, \text{g/mol}} = 0.577 \, \text{mol}
]

Moles of fluorine:

[
n_{\text{F}_2} = \frac{3.08 \, \text{g}}{38.0 \, \text{g/mol}} = 0.0813 \, \text{mol}
]

Step 2: Use the ideal gas law to find the pressures.

We will now use the ideal gas law, ( PV = nRT ), where:

( P ) is pressure,
( V ) is volume,
( n ) is the number of moles,
( R ) is the gas constant (0.0821 L·atm/(mol·K)), and
( T ) is the temperature in Kelvin.

First, assume the temperature ( T ) is constant for both gases. Since both gases are in separate flasks before the stopcock is opened, we can treat them independently.

Step 3: Calculate the initial pressures in each flask.

The total volume of the helium is 2.01 L, and the number of moles is 0.577 mol. Using the ideal gas law:

[
P_{\text{He}} = \frac{n_{\text{He}} RT}{V_{\text{He}}}
]

For the fluorine gas, the volume is 3.16 L, and the number of moles is 0.0813 mol:

[
P_{\text{F}2} = \frac{n{\text{F}2} RT}{V{\text{F}_2}}
]

Step 4: Total pressure after opening the stopcock.

After the stopcock is opened, the total volume of the system is ( V_{\text{total}} = 2.01 + 3.16 = 5.17 \, \text{L} ). The total number of moles is ( n_{\text{total}} = 0.577 + 0.0813 = 0.658 \, \text{mol} ).

The total pressure after the stopcock is opened is given as 625 mm Hg. This can be converted to atm:

[
P_{\text{total}} = 625 \, \text{mm Hg} \times \frac{1 \, \text{atm}}{760 \, \text{mm Hg}} = 0.8237 \, \text{atm}
]

Using the ideal gas law for the total system:

[
P_{\text{total}} = \frac{n_{\text{total}} RT}{V_{\text{total}}}
]

Substitute in the known values:

[
0.8237 \, \text{atm} = \frac{0.658 \, \text{mol} \times 0.0821 \, \text{L·atm/mol·K} \times T}{5.17 \, \text{L}}
]

We can solve for ( T ) (temperature) from this equation.

Step 5: Use the temperature to calculate initial pressure of helium.

Now that we know the temperature ( T ), we can find the initial pressure of helium before the stopcock was opened.

The answer for the initial pressure of helium before the stopcock is opened is calculated as: