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.
Use PV = nRT for both.
Post your work of I can help you through it if you have problems.
Please don't confuse us with different screen names. n = grams/molar mass. R = 0.082057 L*atm/mol*K and I usually round that to 0.08206.
Okay.
PV = nRT
ATM * 2.3L = n (0.08205)(282k)
How do I do “n”?
Okay.
PV = nRT
ATM * 2.3L = n (0.08205)(282k)
How do I do “n”?
A) To calculate the gas pressure inside the tank, we can use the Ideal Gas Law equation, which states:
PV = nRT
Where:
P = Pressure (in units of atm)
V = Volume (in units of L)
n = Number of moles
R = Ideal Gas Constant (0.0821 L.atm/mol.K)
T = Temperature (in units of Kelvin)
First, we need to convert the temperature from Celsius to Kelvin:
9 Celsius + 273 = 282 K
Next, we need to calculate the number of moles of oxygen:
n = mass / molar mass
The molar mass of oxygen (O2) is approximately 32 g/mol.
So, n = 0.29 kg / (32 g/mol) = 9.0625 mol
Now, we can substitute these values into the Ideal Gas Law equation:
PV = nRT
P * 2.3 L = (9.0625 mol) * (0.0821 L.atm/mol.K) * (282 K)
P * 2.3 L = 0.1936 mol.atm
Now we can solve for P:
P = (0.1936 mol.atm) / 2.3 L = 0.0842 atm ≈ 0.08 atm
Rounding off the answer, we get 0.08 atm, which is approximately 0.1 atm.
B) To find the new volume at a different temperature and pressure, we can use the combined gas law equation, which is derived from the ideal gas law:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = Initial Pressure (0.1 atm)
V1 = Initial Volume (2.3L)
T1 = Initial Temperature (282 K)
P2 = Final Pressure (0.95 atm)
V2 = Final Volume (unknown)
T2 = Final Temperature (26 Celsius + 273 = 299 K)
Now we can substitute these values into the combined gas law equation to solve for V2:
(0.1 atm * 2.3 L) / 282 K = (0.95 atm * V2) / 299 K
(0.23 atm.L) / 282 K = (0.95 atm * V2) / 299 K
Cross-multiplying and simplifying, we get:
0.23 * 299 K = 0.95 atm * V2
68.77 = 0.95 V2
Now we solve for V2:
V2 = 68.77 / 0.95 ≈ 72.39 L
Rounding off the answer, we get approximately 72.4 L.
Sure, let's break down the steps to calculate the gas pressure inside the tank and the volume of oxygen at a different temperature and pressure.
A) To calculate the gas pressure inside the tank at 9°C, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
First, we need to convert the given temperature from degrees Celsius to Kelvin. To do that, we add 273 to the Celsius temperature:
9°C + 273 = 282 K
Next, let's calculate the number of moles (n) of oxygen in the tank. We can use the molar mass of O2, which is approximately 32 g/mol:
0.29 kg × (1000 g / 1 kg) × (1 mol / 32 g) = 9.0625 mol
Now, we can substitute the values into the Ideal Gas Law equation:
(P)(2.3 L) = (9.0625 mol)(0.0821 L·atm/mol·K)(282 K)
Simplifying the equation:
P = (9.0625 × 0.0821 × 282) / 2.3
P = 91 atm
Therefore, the gas pressure inside the tank is 91 atm.
B) To calculate the volume of oxygen at 26°C and 0.95 atm, we can use the combined gas law equation:
(P1)(V1) / T1 = (P2)(V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Given values:
P1 = 91 atm
V1 = 2.3 L
T1 = 9°C + 273 = 282 K
P2 = 0.95 atm
T2 = 26°C + 273 = 299 K
Substituting the values into the combined gas law equation:
(91 atm)(2.3 L) / 282 K = (0.95 atm)(V2) / 299 K
Simplifying the equation:
2.3 / 282 = 0.95 V2 / 299
Cross multiplying and solving for V2:
V2 = (2.3 × 0.95 × 299) / 282
V2 ≈ 2.3 × 10^2 L
Therefore, the volume the oxygen would occupy at 26°C and 0.95 atm would be approximately 2.3 × 10^2 L.