The vapor pressure of mercury at 77¢K is 0.0018 mmHg. The atomic weight of mercury is 200.6 g/mole. If the mercury vapor were allowed to reach equilibrium at this temperature in an enclosed space, what would be the mercury vapor concentration in the space?
The problem doesn't give a volume and that affects concn; however, I would use 1 L for V and use PV = nRT, solve for n and convert to grams. That will will you the concn in g/L.
Use PV = nRT
To determine the mercury vapor concentration at equilibrium, we need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
In this case, we have the pressure (P) as 0.0018 mmHg and the temperature (T) as 77 K. Since we want to find the concentration, we need to rearrange the ideal gas law equation to solve for n:
n = PV / RT
Now let's calculate the concentration step by step:
Step 1: Convert the pressure from mmHg to atm.
Since the ideal gas law constant (R) is given in atm, we need to convert the pressure from mmHg to atm.
1 atm = 760 mmHg
So, 0.0018 mmHg * (1 atm / 760 mmHg) = 0.00000236842 atm
Step 2: Convert the atomic weight of mercury to grams per mole.
The atomic weight of mercury is given as 200.6 g/mole.
Step 3: Calculate the concentration using the ideal gas law equation:
n = (0.00000236842 atm) * (V / (0.0821 L·atm/mol·K)) / (77 K)
Here, we use the ideal gas constant R, which is 0.0821 L·atm/mol·K.
Step 4: Simplify the equation by canceling units:
n = (0.00000236842 atm) * (V / 6.7) / 77
Simplifying further:
n = 0.00000035447 (V / 6.7)
So, the mercury vapor concentration in the enclosed space will be 0.00000035447 (V / 6.7) moles per liter or M (molar concentration).