A sample of Chlorine occupies 8.50 L at 80.0 C and 740 mm Hg. What volume will Chlorine occupy at STP?
6.75 L
To calculate the volume of chlorine at STP (Standard Temperature and Pressure), the given conditions need to be converted to STP conditions, which are defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (760 mm Hg).
First, let's convert the temperature from Celsius to Kelvin:
80.0°C + 273.15 = 353.15 K
Next, we can apply the ideal gas law equation, which states:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure (in this case, 740 mm Hg)
V1 = initial volume (in this case, 8.50 L)
T1 = initial temperature (in this case, 353.15 K)
P2 = final pressure (at STP, 760 mm Hg)
V2 = final volume (to be determined)
T2 = final temperature (STP, 273.15 K)
Now we can substitute the values into the equation:
(740 mm Hg * 8.50 L) / 353.15 K = (760 mm Hg * V2) / 273.15 K
Simplifying the equation, we get:
[(740 mm Hg * 8.50 L * 273.15 K)] / [(353.15 K * 760 mm Hg)] = V2
Now, calculate the volume:
V2 = [(740 mm Hg * 8.50 L * 273.15 K)] / [(353.15 K * 760 mm Hg)]
V2 ≈ 7.13 L
Therefore, the volume of chlorine at STP would be approximately 7.13 L.
To find the volume of chlorine at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin:
Given T = 80.0 °C
Convert to Kelvin: T(K) = T(°C) + 273.15
T(K) = 80.0 °C + 273.15 = 353.15 K
Next, we can rearrange the ideal gas law equation to solve for the final volume at STP.
Since we are comparing the initial volume (V1) to the final volume at STP (V2), we can write the equation as:
(V1 / T1) x P1 = (V2 / T2) x P2
Substituting the given values:
(V1 = 8.50 L, T1 = 353.15 K, P1 = 740 mm Hg)
(V2 = ?, T2 = 273.15 K, P2 = 1 atm)
(V2 / 273.15 K) x 1 atm = (8.50 L / 353.15 K) x (740 mm Hg / 1 atm)
To simplify, we need to convert mm Hg to atm:
1 atm = 760 mm Hg
(V2 / 273.15 K) = (8.50 L / 353.15 K) x (740 mm Hg / 760 mm Hg)
(V2 / 273.15 K) = (8.50 L / 353.15 K) x (0.974)
V2 / 273.15 K = 0.02345
Now, we can solve for V2:
V2 = 0.02345 x 273.15 K
V2 ≈ 6.408 L
Therefore, the volume of chlorine at STP is approximately 6.408 liters.