many gases are shipped in high pressure containers. consider a steel tank whose volume is 42.0L and which contains O2 gas at a pressure of 165atm at 23 degrees celsius.
1.) what mass of O2 does the tank contain?
2.) what volume would the gas occupy at STP?
3.) At what temperature would the pressure be 150 atm?
4.) what would be the pressure of the gas if it were cooled to -15 degrees Celsius and put in a container whose volume is 55.0L?
R=0.08206L.atm/mol.K
You can work all of these with one or more of the following:
PV = nRT
n = grams/molar mass
(P1V1/T1) = (P2V2/T2
To solve these questions, we will use the Ideal Gas Law equation, which is given as:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = Ideal Gas Constant
T = temperature
1.) To find the mass of O2 in the tank, we need to calculate the number of moles of O2 first. The molar mass of O2 is 32 g/mol.
n = PV / RT
= (165 atm) * (42.0 L) / (0.08206 L.atm/mol.K * (273.15 + 23) K)
= 30.96 mol
Now, we can calculate the mass of O2:
mass = n * molar mass
= (30.96 mol) * (32 g/mol)
= 990.72 g
Therefore, the tank contains 990.72 grams of O2.
2.) To find the volume of the gas at STP (Standard Temperature and Pressure), we can use the equation:
P1V1 / T1 = P2V2 / T2
To simplify, we can assume that the gas is behaving ideally.
P1 = 165 atm
V1 = 42.0 L
T1 = 23°C = 23 + 273.15 = 296.15 K
P2 = 1 atm (STP)
V2 = ?
T2 = 273.15 K (STP)
Using the formula and substituting the values:
(165 atm) * (42.0 L) / (296.15 K) = (1 atm) * V2 / (273.15 K)
V2 = [(165 atm) * (42.0 L) / (296.15 K)] * (273.15 K) / (1 atm)
= 39.03 L
Therefore, at STP, the gas would occupy approximately 39.03 Liters.
3.) To find the temperature at which the pressure would be 150 atm, we can rearrange the Ideal Gas Law equation:
T = P * V / (n * R)
P = 150 atm
V = 42.0 L
n = 30.96 mol
R = 0.08206 L.atm/mol.K
Substituting the values:
T = (150 atm) * (42.0 L) / (30.96 mol * 0.08206 L.atm/mol.K)
= 256.94 K
Therefore, the temperature at which the pressure would be 150 atm is approximately 256.94 K.
4.) To find the pressure of the gas when cooled to -15°C and put into a container with a volume of 55.0 L, we can use the rearranged Ideal Gas Law equation:
P = n * R * T / V
T = -15°C = -15 + 273.15 = 258.15 K
V = 55.0 L
n = 30.96 mol
R = 0.08206 L.atm/mol.K
Substituting the values:
P = (30.96 mol) * (0.08206 L.atm/mol.K) * (258.15 K) / (55.0 L)
= 121.90 atm
Therefore, the pressure of the gas when cooled to -15°C and put into a container with a volume of 55.0 L would be approximately 121.90 atm.
To answer the questions, we will use the Ideal Gas Law equation: 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.
1.) To find the mass of O2 in the tank, we need to know the number of moles of O2. We can calculate this using the Ideal Gas Law equation. Rearranging the equation to solve for n, we have n = PV / RT.
Given:
P = 165 atm
V = 42.0 L
T = 23°C = 23 + 273 = 296 K (remember to convert Celsius to Kelvin)
R = 0.08206 L.atm/mol.K
Substituting the values into the equation, we get:
n = (165 atm * 42.0 L) / (0.08206 L.atm/mol.K * 296 K)
Solving this equation, we find n = 27.24 moles (rounded to two decimal places).
To find the mass of O2, we need to know the molar mass of O2, which is approximately 32 g/mol. So, the mass of O2 in the tank is:
Mass = n * molar mass = 27.24 moles * 32 g/mol = 872.32 grams (rounded to two decimal places).
Therefore, the tank contains approximately 872.32 grams of O2.
2.) To find the volume the gas would occupy at Standard Temperature and Pressure (STP), we can use the Ideal Gas Law equation again. At STP, the pressure is 1 atm and the temperature is 273 K.
Given:
P = 165 atm
V = 42.0 L
T = 23°C = 23 + 273 = 296 K
R = 0.08206 L.atm/mol.K
We need to find the new volume, V_STP. Rearranging the Ideal Gas Law equation, we have:
V_STP = (n * R * T_STP) / P_STP
Substituting the values into the equation, we get:
V_STP = (27.24 moles * 0.08206 L.atm/mol.K * 273 K) / (1 atm)
Solving this equation, we find V_STP = 616.44 L (rounded to two decimal places).
Therefore, the gas would occupy approximately 616.44 liters at STP.
3.) To find the temperature at which the pressure would be 150 atm, we need to rearrange the Ideal Gas Law equation to solve for T.
Given:
P = 165 atm
V = 42.0 L
T = unknown
R = 0.08206 L.atm/mol.K
We need to find the temperature, T_new. Rearranging the equation, we have:
T_new = (P_new * V) / (n * R)
Substituting the values into the equation, we get:
T_new = (150 atm * 42.0 L) / (27.24 moles * 0.08206 L.atm/mol.K)
Solving this equation, we find T_new = 275.65 K (rounded to two decimal places).
Therefore, the temperature at which the pressure would be 150 atm is approximately 275.65 Kelvin.
4.) To find the pressure of the gas when cooled to -15 degrees Celsius and placed in a container with a volume of 55.0 L, we will apply the Ideal Gas Law equation.
Given:
P = unknown
V = 55.0 L
T = -15°C = -15 + 273 = 258 K
R = 0.08206 L.atm/mol.K
We need to find the pressure, P_new. Rearranging the equation, we have:
P_new = (n * R * T_new) / V_new
Substituting the values into the equation, we get:
P_new = (27.24 moles * 0.08206 L.atm/mol.K * 258 K) / (55.0 L)
Solving this equation, we find P_new = 32.77 atm (rounded to two decimal places).
Therefore, the pressure of the gas when cooled to -15 degrees Celsius and placed in a container with a volume of 55.0 L would be approximately 32.77 atm.
Note: Make sure to convert all temperatures to Kelvin, as the Ideal Gas Law requires temperature in Kelvin.