A 22.0-L tank of carbon dioxide gas (CO2) is at a pressure of 9.40 105 Pa and temperature of 22.0°C.
(a) Calculate the temperature of the gas in Kelvin.
K
(b) Use the ideal gas law to calculate the number of moles of gas in the tank.
mol
(c) Use the periodic table to compute the molecular weight of carbon dioxide, expressing it in grams per mole.
g/mol
(d) Obtain the number of grams of carbon dioxide in the tank.
g
(e) A fire breaks out, raising the ambient temperature by 224.0 K while 82.0 g of gas leak out of the tank. Calculate the new temperature and the number of moles of gas remaining in the tank.
temperature
K
number of moles
mol
(f) Using the ideal gas law, find a symbolic expression for the final pressure, neglecting the change in volume of the tank. (Use the following as necessary: ni, the initial number of moles; nf, the final number of moles; Ti, the initial temperature; Tf, the final temperature; and Pi, the initial pressure.)
Pf
=
(g) Calculate the final pressure in the tank as a result of the fire and leakage.
Pa
(a) To convert the temperature from Celsius to Kelvin, add 273.15:
T(K) = T(°C) + 273.15
T(K) = 22.0 + 273.15
T(K) = 295.15 K
(b) Use the ideal gas law:
PV = nRT
Rearrange for n:
n = PV/(RT)
Given values are:
P = 9.40 * 10^5 Pa
V = 22.0 L = 0.022 m^3 (convert L to m^3 by multiplying with 0.001)
T = 295.15 K
R = 8.314 J/(mol*K) (gas constant)
n = (9.40 * 10^5 * 0.022) / (8.314 * 295.15)
n = 9.66 mol
(c) The molecular weight of carbon dioxide (CO2) is the sum of the atomic weights of carbon and oxygen:
CO2 = C + 2O
M(CO2) = 12.01 + 2 * 16.00
M(CO2) = 12.01 + 32.00
M(CO2) = 44.01 g/mol
(d) To obtain the number of grams, multiply the molecular weight by the number of moles:
mass = M(CO2) * n
mass = 44.01 * 9.66
mass = 425.05 g
(e) First, calculate the new temperature:
Tf = Ti + 224.0
Tf = 295.15 + 224.0
Tf = 519.15 K
To find the new number of moles, subtract the leaked gas in grams (82.0 g) from the initial mass (425.05 g) and divide by the molecular weight:
initial mass = 425.05 g
leaked gas = 82.0 g
remaining mass = 425.05 - 82.0 = 343.05 g
nf = remaining mass / M(CO2)
nf = 343.05 / 44.01
nf = 7.79 mol
(f) According to the ideal gas law and considering constant volume:
Pf/niTi = nfTf/Pi
Rearrange for Pf:
Pf = nfTfTi / (niTi)
(g) Calculate the final pressure:
Pf = (7.79 * 519.15) / (9.66 * 295.15)
Pf = 4028178.85 / 2852.764
Pf = 1412590.41 Pa
(a) To convert the temperature from Celsius to Kelvin, we use the formula:
T(K) = T(°C) + 273.15
In this case, T(°C) = 22.0°C.
T(K) = 22.0°C + 273.15 = 295.15 K
Therefore, the temperature of the gas is 295.15 K.
(b) We can use the ideal gas law equation to calculate the number of moles:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
R = 8.314 J/(mol·K) (gas constant)
Rearranging the equation to solve for n:
n = (PV) / (RT)
Given:
P = 9.40 x 10^5 Pa (pressure)
V = 22.0 L (volume)
T = 295.15 K (temperature)
Substituting the values:
n = (9.40 x 10^5 Pa * 22.0 L) / (8.314 J/(mol·K) * 295.15 K)
Simplifying:
n ≈ 8.61 moles
Therefore, the number of moles of gas in the tank is approximately 8.61 mol.
(c) The molecular weight of carbon dioxide (CO2) can be found from the periodic table. Carbon (C) has a molar mass of 12.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Since carbon dioxide contains one carbon atom and two oxygen atoms, we calculate the molecular weight as follows:
Molecular weight of CO2 = 12.01 g/mol + 2 * 16.00 g/mol
Molecular weight of CO2 = 12.01 g/mol + 32.00 g/mol
Molecular weight of CO2 = 44.01 g/mol
Therefore, the molecular weight of carbon dioxide is 44.01 g/mol.
(d) To find the mass of carbon dioxide in the tank, we can use the formula:
mass = number of moles * molecular weight
Given:
Number of moles = 8.61 mol
Molecular weight = 44.01 g/mol
mass = 8.61 mol * 44.01 g/mol
mass ≈ 378.87 g
Therefore, the number of grams of carbon dioxide in the tank is approximately 378.87 g.
(e) After the fire and leakage, the ambient temperature increases by 224.0 K, and 82.0 g of gas leaks out. We need to calculate the new temperature and the number of moles of gas remaining in the tank.
First, we can calculate the new temperature:
New temperature = initial temperature + temperature increase
New temperature = 295.15 K + 224.0 K
New temperature = 519.15 K
Next, we can calculate the moles of gas remaining using the formula:
moles of gas remaining = initial moles - moles leaked
Given:
initial moles = 8.61 mol
moles leaked = 82.0 g / molecular weight
moles leaked = 82.0 g / 44.01 g/mol
moles leaked ≈ 1.86 mol
moles of gas remaining = 8.61 mol - 1.86 mol
moles of gas remaining ≈ 6.75 mol
Therefore, the new temperature is approximately 519.15 K and the number of moles of gas remaining in the tank is approximately 6.75 mol.
(f) Using the ideal gas law, we can find a symbolic expression for the final pressure, neglecting the change in volume of the tank. The ideal gas law equation is:
PV = nRT
Rearranging the equation:
P = (nR)/V * T
Let's define the initial conditions as:
ni = initial number of moles
Ti = initial temperature
Pi = initial pressure
Therefore, the expression for the final pressure neglecting the change in volume is:
Pf = (ni * R * Tf) / (ni * Ti)
(g) To calculate the final pressure, we substitute the known values into the equation:
Given:
ni = 8.61 mol (initial number of moles)
Ti = 295.15 K (initial temperature)
Tf = 519.15 K (final temperature)
Pf = (8.61 mol * 8.314 J/(mol·K) * 519.15 K) / (8.61 mol * 295.15 K)
Pf ≈ 22,428.67 Pa
Therefore, the final pressure in the tank, as a result of the fire and leakage, is approximately 22,428.67 Pa.
To solve this problem, we will use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
(a) To convert the temperature from degrees Celsius to Kelvin, we use the formula:
T(K) = T(°C) + 273.15.
So, we have:
T(K) = 22.0°C + 273.15 = 295.15 K
(b) The ideal gas law can be rearranged to solve for the number of moles:
n = PV / RT.
Given:
P = 9.40 × 10^5 Pa
V = 22.0 L
R = 8.314 J/(mol·K)
T = 295.15 K
Plugging in the values, we get:
n = (9.40 × 10^5 Pa) × (22.0 L) / (8.314 J/(mol·K) × 295.15 K)
Simplifying the units, we get the number of moles:
n = (9.40 × 10^5 Pa) × (22.0 L) / (8.314 J·L/(mol·K) × 295.15 K)
n ≈ 878.98 mol
(c) To compute the molecular weight of carbon dioxide (CO2), we need to find the atomic masses of carbon (C) and oxygen (O) from the periodic table. The atomic mass of carbon is approximately 12.01 g/mol, and the atomic mass of oxygen is approximately 16.00 g/mol.
So, the molecular weight of CO2 is:
Molecular weight of CO2 = (12.01 g/mol) + 2 × (16.00 g/mol)
Molecular weight of CO2 ≈ 44.01 g/mol
(d) To obtain the number of grams of carbon dioxide in the tank, we use the formula:
Mass = number of moles × molecular weight.
Given:
Number of moles (n) = 878.98 mol
Molecular weight of carbon dioxide (CO2) = 44.01 g/mol
Plugging in the values, we get:
Mass = 878.98 mol × 44.01 g/mol
Mass ≈ 38,668.37 g or 38.67 kg
(e) Given:
Change in temperature (ΔT) = 224.0 K
Mass of gas leaked (m) = 82.0 g
The new temperature (Tf) can be calculated as:
Tf = Ti + ΔT
Tf = 295.15 K + 224.0 K
Tf = 519.15 K
To calculate the remaining number of moles (nf), we need to consider the mass of gas leaked.
Number of moles of leaked gas (n_leaked) = m / molecular weight of CO2.
Number of moles remaining in the tank (nf) = ni - n_leaked.
Given:
Molecular weight of CO2 = 44.01 g/mol
Plugging in the values, we get:
n_leaked = 82.0 g / 44.01 g/mol
n_leaked ≈ 1.863 mol
nf = 878.98 mol - 1.863 mol
nf ≈ 877.12 mol
(f) The ideal gas law equation can be used to find a symbolic expression for the final pressure (Pf), neglecting the change in volume of the tank:
Pf = (nf / ni) * (Ti / Tf) * Pi
(g) To calculate the final pressure in the tank as a result of the fire and leakage, we need to calculate the values of nf, Ti, and Pi (initial pressure).
Given:
nf = 877.12 mol
Ti = 295.15 K
Pi = 9.40 × 10^5 Pa
Plugging in the values, we get:
Pf = (877.12 mol / ni) * (295.15 K / Tf) * (9.40 × 10^5 Pa)
(Note: The value of ni is not provided in the question. You may need to refer to additional information to obtain this value.)
Solving this equation will give you the final pressure (Pf) in pascals (Pa).