A typical fluorescent light bulb contains argon gas and mercury vapor.?
a. If the total gas pressure in a fluorescent light bulb is 307.1 Pa and the partial pressure of the argon gas is 306.0 Pa, what is the partial pressure of the mercury vapor?
b. If a 935-mL fluorescent bulb contains 4108 mg of argon gas and 77 mg of mercury vapor, what is the total gas pressure in the fluorescent bulb at 313 K?
a. Ptotal = pHg + pAr
b. mols Ar gas = grams/molar mass = ?
mols Hg gas = grams/molar mass = ?
Total mols gas = nHg + nAr
Then use PV = nRT and solve for P at the conditions listed. Remember V is in L.
a. To find the partial pressure of the mercury vapor in the fluorescent light bulb, we can use Dalton's Law of partial pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each gas component.
Given:
Total gas pressure (P_total) = 307.1 Pa
Partial pressure of argon (P_argon) = 306.0 Pa
To find the partial pressure of mercury vapor (P_mercury), we can subtract the partial pressure of argon from the total gas pressure:
P_mercury = P_total - P_argon
P_mercury = 307.1 Pa - 306.0 Pa
P_mercury = 1.1 Pa
Therefore, the partial pressure of the mercury vapor in the fluorescent light bulb is 1.1 Pa.
b. To find the total gas pressure in the fluorescent bulb, we need to calculate the individual partial pressures of argon and mercury vapor. Then, we can add these partial pressures together using Dalton's Law.
Given:
Volume of fluorescent bulb (V) = 935 mL = 935 x 10^(-6) m^3
Amount of argon gas (n_argon) = 4108 mg = 4108 x 10^(-6) kg
Amount of mercury vapor (n_mercury) = 77 mg = 77 x 10^(-6) kg
Temperature (T) = 313 K
First, let's calculate the partial pressure of argon:
Using the ideal gas law, we can relate the amount of a gas (in moles) to its partial pressure:
P_argon = (n_argon * R * T) / V
Where:
R is the ideal gas constant (8.314 J/(mol·K))
P_argon = (4108 x 10^(-6) kg * 8.314 J/(mol·K) * 313 K) / 935 x 10^(-6) m^3
P_argon = 0.00138009 Pa
Next, let's calculate the partial pressure of mercury vapor:
P_mercury = (n_mercury * R * T) / V
P_mercury = (77 x 10^(-6) kg * 8.314 J/(mol·K) * 313 K) / 935 x 10^(-6) m^3
P_mercury = 0.00004041 Pa
Finally, let's find the total gas pressure:
P_total = P_argon + P_mercury
P_total = 0.00138009 Pa + 0.00004041 Pa
P_total = 0.0014205 Pa
Therefore, the total gas pressure in the fluorescent bulb at 313 K is 0.0014205 Pa.
To solve these problems, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
a. To find the partial pressure of the mercury vapor, we need to subtract the partial pressure of the argon gas from the total gas pressure.
Given:
Total gas pressure (Ptotal) = 307.1 Pa
Partial pressure of argon gas (Pargon) = 306.0 Pa
To find the partial pressure of mercury vapor (Pmercury), we can use the equation:
Pmercury = Ptotal - Pargon
Pmercury = 307.1 Pa - 306.0 Pa
Pmercury = 1.1 Pa
Therefore, the partial pressure of the mercury vapor is 1.1 Pa.
b. To find the total gas pressure in the fluorescent bulb, we need to convert the mass of the gases into moles and then apply the ideal gas law equation.
Given:
Volume (V) = 935 mL = 0.935 L
Argon gas mass (Margon) = 4108 mg = 4.108 g
Mercury vapor mass (Mmercury) = 77 mg = 0.077 g
Temperature (T) = 313 K
First, we need to convert the masses into moles:
Number of moles of argon gas (nargon) = Margon / molar mass of argon
Number of moles of mercury vapor (nmercury) = Mmercury / molar mass of mercury
The molar mass of argon (Ar) is approximately 39.95 g/mol, and the molar mass of mercury (Hg) is approximately 200.59 g/mol.
nargon = 4.108 g / 39.95 g/mol
nargon ≈ 0.1029 mol
nmercury = 0.077 g / 200.59 g/mol
nmercury ≈ 0.000384 mol
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
Ptotal * V = (nargon + nmercury) * R * T
To find the total gas pressure (Ptotal), we rearrange the equation:
Ptotal = (nargon + nmercury) * R * T / V
Let's calculate it:
Ptotal = (0.1029 mol + 0.000384 mol) * (8.314 J/mol·K) * 313 K / 0.935 L
Ptotal ≈ 2920.72 J/mol
Therefore, the total gas pressure in the fluorescent bulb at 313 K is approximately 2920.72 Pa.