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.
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.
Sure! Let's calculate the gas pressure inside the tank and the volume of the oxygen at a different temperature and pressure.
A) To calculate the gas pressure inside the tank, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in Pascals or atmospheres)
V = Volume (in liters)
n = moles of gas
R = Ideal gas constant (0.0821 L·atm/mol·K for ideal gases)
T = Temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin. The Kelvin temperature scale is the same as Celsius, but with 273.15 added to it. So, 9 Celsius would be 9 + 273.15 = 282.15 Kelvin.
Next, we need to calculate the number of moles of oxygen in the tank. To do this, we'll use the molar mass of oxygen, which is approximately 32 g/mol.
1. Convert the mass of oxygen from kg to grams:
0.29 kg x 1000 g/kg = 290 g
2. Convert grams of oxygen to moles:
moles = mass / molar mass
moles = 290 g / 32 g/mol ≈ 9.06 mol
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P(2.3 L) = (9.06 mol)(0.0821 L·atm/mol·K)(282.15 K)
Simplifying the equation:
P(2.3) = 233.959 atm·L
P = 233.959 atm·L / 2.3 L ≈ 101.73 atm ≈ 102 atm
Therefore, the gas pressure inside the tank at 9 degrees Celsius is approximately 102 atmospheres.
B) To calculate the new volume of the oxygen at a different temperature and pressure, we can rearrange the ideal gas law equation:
(V1 / T1) = (V2 / T2)
Where:
V1 = Initial Volume
T1 = Initial Temperature (in Kelvin)
V2 = Final Volume
T2 = Final Temperature (in Kelvin)
We are given:
V1 = 2.3 L
T1 = 282.15 K
T2 = 26 + 273.15 = 299.15 K
P2 = 0.95 atm
Rearranging the equation, we can solve for V2:
V2 = (V1 / T1) * T2
Plugging in the values:
V2 = (2.3 L / 282.15 K) * 299.15 K
V2 ≈ 2354.38 L ≈ 2.35 x 10^3 L
Therefore, the volume of the oxygen at 26 degrees Celsius and 0.95 atm is approximately 2.35 x 10^3 L.
I hope this helps! Let me know if you have any further questions.
To solve these problems, we can use the ideal gas law equation, which is given by PV = nRT. This equation relates pressure (P), volume (V), the number of moles of gas (n), the ideal gas constant (R), and temperature (T).
Let's go through each part step by step:
A) To calculate the gas pressure inside the tank at 9 degrees Celsius (or 9 degrees Celsius above absolute zero, which is 9 + 273.15 Kelvin = 282.15K), we need to determine the number of moles of O2 in the tank. We can use the ideal gas law equation to find the number of moles (n).
1. Convert the given mass of oxygen (0.29 kg) to moles:
To do this, we need to know the molar mass of oxygen, which is approximately 32 g/mol.
0.29 kg = 290 g (since there are 1000 grams in a kilogram)
moles = mass / molar mass
moles = 290 g / 32 g/mol ≈ 9.06 mol
2. We can plug the values into the ideal gas law equation:
PV = nRT
Here, we have:
P = gas pressure (we need to find this)
V = volume of the tank (2.3 L)
n = number of moles of gas (9.06 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (282.15 K)
Plugging in these values, we get:
P * 2.3 L = 9.06 mol * 0.0821 L·atm/(mol·K) * 282.15 K
3. Rearranging the equation to isolate P, the pressure:
P = (9.06 mol * 0.0821 L·atm/(mol·K) * 282.15 K) / 2.3 L
P ≈ 91 atm
Therefore, the gas pressure inside the tank at 9 degrees Celsius is approximately 91 atm.
B) To find the volume that this oxygen would occupy at 26 degrees Celsius (or 26 degrees Celsius above absolute zero, which is 26 + 273.15 Kelvin = 299.15K) and a pressure of 0.95 atm, we can again use the ideal gas law equation.
We need to find the new volume of the oxygen, so we rearrange the equation to isolate V:
PV = nRT
Here, we have:
P = gas pressure (0.95 atm)
V = volume of the tank (we need to find this)
n = number of moles of gas (9.06 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (299.15 K)
Plugging in these values, we get:
0.95 atm * V = 9.06 mol * 0.0821 L·atm/(mol·K) * 299.15 K
Rearranging the equation to isolate V, the volume:
V = (9.06 mol * 0.0821 L·atm/(mol·K) * 299.15 K) / 0.95 atm
V ≈ 230 L or 2.3 x 10^2 L
Therefore, the volume this oxygen would occupy at 26 degrees Celsius and 0.95 atm is approximately 2.3 x 10^2 L.