Question
A scuba diver’s tank contains 0.29 kg of O2 compressed into a volume of 2.3L.
A)calculate the gas pressure inside the tank at 9 Celsius.
B) what volume would this oxygen occupy at 26 Celsius and 0.95atm?
Answer is a)91atm
B) 2.3 x 10^2 L
Trying to find out how they got these. Please do it step by step with me.
A)calculate the gas pressure inside the tank at 9 Celsius.
B) what volume would this oxygen occupy at 26 Celsius and 0.95atm?
Answer is a)91atm
B) 2.3 x 10^2 L
Trying to find out how they got these. Please do it step by step with me.
Answers
GPT 3.5
To solve these problems, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) for this problem)
T = temperature in Kelvin
So let's solve A):
Given:
Mass of O2 = 0.29 kg
Volume = 2.3 L
Temperature = 9 Celsius
Step 1: Convert temperature from Celsius to Kelvin.
Kelvin = Celsius + 273.15
T = 9 + 273.15 = 282.15 K
Step 2: Convert mass of O2 to moles.
To do this, we need to know the molar mass of O2, which is 32 g/mol. Since we have the mass in kg, we need to convert it to grams first.
Mass of O2 in grams = 0.29 kg x 1000 g/kg = 290 g
Number of moles = Mass / Molar mass
n = 290 g / 32 g/mol = 9.06 mol
Step 3: Plug the values into the ideal gas law equation to solve for the pressure.
PV = nRT
P * 2.3 = 9.06 * 0.0821 * 282.15
P = (9.06 * 0.0821 * 282.15) / 2.3
P ≈ 91 atm
Therefore, the gas pressure inside the tank at 9 Celsius is approximately 91 atm.
Now let's solve B):
Given:
Volume = 2.3 L
Temperature = 26 Celsius
Pressure = 0.95 atm
Step 1: Convert temperature from Celsius to Kelvin.
T = 26 + 273.15 = 299.15 K
Step 2: Plug the values into the ideal gas law equation to solve for the number of moles.
PV = nRT
(0.95) * V = n * (0.0821) * (299.15)
n = (0.95 * V) / (0.0821 * 299.15)
n ≈ 0.118 mol
Step 3: Use the number of moles to calculate the new volume at the given conditions.
PV = nRT
V = (n * R * T) / P
V = (0.118 * 0.0821 * 299.15) / 0.95
V ≈ 2.3 x 10^2 L
Therefore, the volume that this oxygen would occupy at 26 Celsius and 0.95 atm is approximately 2.3 x 10^2 L.
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) for this problem)
T = temperature in Kelvin
So let's solve A):
Given:
Mass of O2 = 0.29 kg
Volume = 2.3 L
Temperature = 9 Celsius
Step 1: Convert temperature from Celsius to Kelvin.
Kelvin = Celsius + 273.15
T = 9 + 273.15 = 282.15 K
Step 2: Convert mass of O2 to moles.
To do this, we need to know the molar mass of O2, which is 32 g/mol. Since we have the mass in kg, we need to convert it to grams first.
Mass of O2 in grams = 0.29 kg x 1000 g/kg = 290 g
Number of moles = Mass / Molar mass
n = 290 g / 32 g/mol = 9.06 mol
Step 3: Plug the values into the ideal gas law equation to solve for the pressure.
PV = nRT
P * 2.3 = 9.06 * 0.0821 * 282.15
P = (9.06 * 0.0821 * 282.15) / 2.3
P ≈ 91 atm
Therefore, the gas pressure inside the tank at 9 Celsius is approximately 91 atm.
Now let's solve B):
Given:
Volume = 2.3 L
Temperature = 26 Celsius
Pressure = 0.95 atm
Step 1: Convert temperature from Celsius to Kelvin.
T = 26 + 273.15 = 299.15 K
Step 2: Plug the values into the ideal gas law equation to solve for the number of moles.
PV = nRT
(0.95) * V = n * (0.0821) * (299.15)
n = (0.95 * V) / (0.0821 * 299.15)
n ≈ 0.118 mol
Step 3: Use the number of moles to calculate the new volume at the given conditions.
PV = nRT
V = (n * R * T) / P
V = (0.118 * 0.0821 * 299.15) / 0.95
V ≈ 2.3 x 10^2 L
Therefore, the volume that this oxygen would occupy at 26 Celsius and 0.95 atm is approximately 2.3 x 10^2 L.
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