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The vapor pressure of mercury at 79F is 0.0002 mmHg. What is the concentration of Hg in ppm in a 500 ft3 room saturated with mercury vapor at 79F

User MarkHim
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Answer:

The concentration of mercury (Hg) in the room saturated with mercury vapor at 79°F is approximately 5.99 ppm.

Step-by-step explanation:

To find the concentration of mercury (Hg) in parts per million (ppm) in a 500 ft³ room saturated with mercury vapor at 79°F, you can use the ideal gas law and the vapor pressure of mercury.

First, convert the temperature from Fahrenheit (°F) to Kelvin (K):

T(K) = (T(°F) - 32) × 5/9 + 273.15

T(K) = (79°F - 32) × 5/9 + 273.15 ≈ 298.15 K

Now, use the ideal gas law to find the number of moles of mercury vapor in the room:

PV = nRT

Where:

P = Pressure (vapor pressure of mercury) = 0.0002 mmHg (convert to atm)

V = Volume = 500 ft³ (convert to liters)

n = Number of moles

R = Ideal gas constant = 0.0821 L·atm/mol·K

T = Temperature in Kelvin (298.15 K)

Convert pressure from mmHg to atm:

1 mmHg = 0.00131579 atm

0.0002 mmHg = 0.0002 × 0.00131579 atm ≈ 2.63158 × 10^-7 atm

Now, plug in these values:

(2.63158 × 10^-7 atm) * (V = 500 ft³ * 28.3168 L/ft³) = n * (0.0821 L·atm/mol·K) * (298.15 K)

Solve for n (number of moles):

n ≈ (2.63158 × 10^-7 atm * 500 ft³ * 28.3168 L/ft³) / (0.0821 L·atm/mol·K * 298.15 K)

n ≈ 0.000003377 moles

Now that you have the number of moles of mercury vapor in the room, you can calculate the concentration in ppm.

1 ppm = 1 part per million, which is equivalent to 1 mole of solute per 1 million moles of solution.

Concentration (ppm) = (moles of solute / moles of solution) * 1,000,000

Concentration (ppm) = (0.000003377 moles / (0.000003377 moles + moles of air in the room)) * 1,000,000

Since air is mostly composed of nitrogen and oxygen and we assume it behaves ideally, you can use the ideal gas law to estimate the moles of air in the room. At standard conditions (1 atm and 298.15 K), 1 mole of any ideal gas occupies approximately 22.4 liters. Therefore:

moles of air = (P * V) / (R * T)

moles of air = (1 atm * 500 ft³ * 28.3168 L/ft³) / (0.0821 L·atm/mol·K * 298.15 K)

Now, calculate the concentration:

Concentration (ppm) = (0.000003377 moles / (0.000003377 moles + moles of air)) * 1,000,000

Calculate moles of air first:

moles of air ≈ (1 atm * 500 ft³ * 28.3168 L/ft³) / (0.0821 L·atm/mol·K * 298.15 K)

moles of air ≈ 564.49 moles

Now calculate the concentration:

Concentration (ppm) = (0.000003377 moles / (0.000003377 moles + 564.49 moles)) * 1,000,000

Concentration (ppm) ≈ (0.000003377 / (0.000003377 + 564.49)) * 1,000,000 ≈ 5.99 ppm

So, the concentration of mercury (Hg) in the room saturated with mercury vapor at 79°F is approximately 5.99 ppm.

User Peter Weyand
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