A tank contains 12.2 g of chlorine gas (Cl2) at a temperature of 84 °C and an absolute pressure of 5.60 × 105 Pa. The mass per mole of Cl2 is 70.9 g/mol. (a) Determine the volume of the tank. (b) Later, the temperature of the tank has dropped to 26 °C and, due to a leak, the pressure has dropped to 3.20 × 105 Pa. How many grams of chlorine gas have leaked out of the tank?
how do you solve this???
To solve these problems, we will use the ideal gas law:
PV = nRT
where:
- P is the pressure (in Pa)
- V is the volume (in m^3)
- n is the number of moles
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature (in Kelvin)
First, let's solve part (a) and find the volume of the tank.
Step 1: Convert the temperature from Celsius to Kelvin.
T1 = 84 °C + 273.15 = 357.15 K
Step 2: Convert the pressure to Pascals.
P1 = 5.60 × 10^5 Pa
Step 3: Calculate the number of moles of chlorine gas.
n = mass / molar mass
n = 12.2 g / 70.9 g/mol ≈ 0.172 mol
Step 4: Plug the values into the ideal gas law and solve for volume (V).
PV = nRT
V = nRT / P1
V = (0.172 mol)(8.314 J/(mol·K))(357.15 K) / (5.60 × 10^5 Pa)
V ≈ 0.0122 m^3
Therefore, the volume of the tank is approximately 0.0122 m^3.
Now, let's solve part (b) to find the amount of chlorine gas leaked.
Step 1: Convert the new temperature from Celsius to Kelvin.
T2 = 26 °C + 273.15 = 299.15 K
Step 2: Convert the new pressure to Pascals.
P2 = 3.20 × 10^5 Pa
Step 3: Calculate the new volume of the tank at the lower temperature and pressure, assuming the amount of gas remaining in the tank is the same as before the leak (i.e., the number of moles remains constant).
V2 = nRT2 / P2
V2 = (0.172 mol)(8.314 J/(mol·K))(299.15 K) / (3.20 × 10^5 Pa)
V2 ≈ 0.0140 m^3
Step 4: Calculate the amount of gas leaked by finding the difference in volume.
Volume leaked = V1 - V2
Volume leaked ≈ 0.0122 m^3 - 0.0140 m^3
Volume leaked ≈ -0.0018 m^3
Since volume cannot be negative, the negative sign indicates that something was taken away. Therefore, the absolute value represents the volume leaked. So, the volume leaked is approximately 0.0018 m^3.
Step 5: Calculate the mass of the leaked chlorine gas.
Mass of leaked gas = volume leaked x density
Mass of leaked gas = volume leaked x molar mass / molar volume
Mass of leaked gas = (0.0018 m^3) x (70.9 g/mol) / (0.0224 m^3)
Mass of leaked gas ≈ 5.727 g
Therefore, approximately 5.727 grams of chlorine gas have leaked out of the tank.
To solve this problem, we'll need to use the ideal gas law, which states that the pressure of a gas is proportional to its temperature, volume, and the number of moles of the gas.
The ideal gas law equation is: PV = nRT
Where:
P = pressure (in Pa)
V = volume (in m^3)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, let's solve part (a) to determine the volume of the tank. We have the following information:
Mass of chlorine gas (m) = 12.2 g
Mass per mole of Cl2 = 70.9 g/mol
Temperature (T) = 84 °C = 84 + 273.15 = 357.15 K
Absolute pressure (P) = 5.60 × 10^5 Pa
Ideal gas constant (R) = 8.314 J/(mol·K)
To find the number of moles (n) of chlorine gas in the tank, we'll use the formula: n = m / M, where M is the molar mass of chlorine gas.
Molar mass of Cl2 = 2 * 35.45 g/mol = 70.9 g/mol
n = 12.2 g / 70.9 g/mol = 0.172 mol
Now, we can rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
Substituting the known values, we get:
V = (0.172 mol * 8.314 J/(mol·K) * 357.15 K) / (5.60 × 10^5 Pa)
Calculating this, we find:
V ≈ 0.018 m^3
Therefore, the volume of the tank is approximately 0.018 m^3.
Moving on to part (b), we need to determine the amount of chlorine gas that has leaked out of the tank.
The new temperature (T') = 26 °C = 26 + 273.15 = 299.15 K
The new pressure (P') = 3.20 × 10^5 Pa
From the ideal gas law equation PV = nRT, we can rearrange it to calculate the new number of moles (n'):
n' = (P' * V) / (R * T')
Substituting the known values, we have:
n' = (3.20 × 10^5 Pa * 0.018 m^3) / (8.314 J/(mol·K) * 299.15 K)
Calculating this, we find:
n' ≈ 0.242 mol
The moles of chlorine gas that leaked out of the tank is the difference between the initial and final number of moles:
Moles leaked (Δn) = n - n'
Δn = 0.172 mol - 0.242 mol
Calculating this, we find:
Δn ≈ -0.070 mol
Since moles cannot be negative, we assume that the negative sign indicates a loss of 0.070 moles of chlorine gas.
To determine the mass of chlorine gas leaked out, we multiply the moles leaked by the molar mass of Cl2:
Mass of leaked chlorine gas = Δn * M
Mass of leaked chlorine gas = -0.070 mol * 70.9 g/mol
Calculating this, we find:
Mass of leaked chlorine gas ≈ -4.963 g
Since mass cannot be negative, we assume that 4.963 g of chlorine gas have leaked out of the tank.
Therefore, the answer to part (b) is approximately 4.963 g.