What volume (in liters) is occupied by 3.85×10^22 nitrogen molecules at 105 C and 255 mmHg?
Use PV = nRT.
n = (3.85E22/6.02E23)
P = 255/760 = ? atm
R = 0.0806 L*atm/mol*K
T = 273 + 105 = ?K
Solve for V in L.
To find the volume occupied by the nitrogen molecules, we can use the Ideal Gas Law equation, which states:
PV = nRT
Where:
P = pressure (in units of force per unit area, such as Pa or atm)
V = volume (in units of cubic meters or liters)
n = number of moles
R = ideal gas constant (in units that depend on the units used for pressure, volume, and temperature)
T = temperature (in units of Kelvin)
Rearranging the equation to solve for volume (V), we get:
V = (nRT) / P
Now, let's plug in the known values:
n = number of molecules / Avogadro's number
R = ideal gas constant
T = temperature in Kelvin
P = pressure in atm
Avogadro's number is approximately 6.022 x 10^23 molecules/mol, so 3.85 x 10^22 molecules is equivalent to:
(3.85 x 10^22) / (6.022 x 10^23) = 0.064 mol of nitrogen molecules
The ideal gas constant, R, is approximately 0.0821 L·atm/(mol·K).
Now, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 105 + 273.15 = 378.15 K
Finally, substitute the values into the equation:
V = (0.064 mol * 0.0821 L·atm/(mol·K) * 378.15 K) / (255 mmHg / 760 mmHg/atm)
Simplifying the expression, we have:
V = (0.064 * 0.0821 * 378.15) / (255 / 760) liters
Calculating the expression, we find:
V ≈ 0.406 liters
Therefore, the volume occupied by 3.85×10^22 nitrogen molecules at 105°C and 255 mmHg is approximately 0.406 liters.