A scuba diver's tank contains .25 kg of oxygen compressed into a volume of 4.3 L (for oxygen, 32g = 1 mole)
Calculate the gas pressure inside the tank at 9 degrees Celsius
What volume would this occupy at 26 degrees Celsius and .95 atmospheres?
Use the ideal gas law. moles O2= 250/molmassO2
P= nRT/V
what would the number of moles equal then? and R= .0821 K atm L mol?
I still don't get it.. lol
.29kg=290g*32g(Molar mass of O2)
To calculate the gas pressure inside the tank at 9 degrees Celsius, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is:
T (Kelvin) = T (Celsius) + 273.15
So, for 9 degrees Celsius:
T = 9 + 273.15 = 282.15 K
Next, we need to calculate the number of moles (n) of oxygen present in the tank. We can use the molar mass of oxygen to convert the mass of oxygen in the tank to moles:
Molar mass of Oxygen = 32 g/mol
Mass of oxygen in the tank = 0.25 kg = 250 g
Number of moles (n) = Mass / Molar mass
= 250 g / 32 g/mol
= 7.8125 moles
Now, we can use the ideal gas law equation to calculate the pressure (P):
PV = nRT
P * 4.3 L = 7.8125 moles * 0.0821 L·atm/mol·K * 282.15 K
P * 4.3 L = 7.8125 moles * 23.845525 L·atm/mol
P = (7.8125 moles * 23.845525 L·atm/mol) / 4.3 L
P = 43.54076 atm
Therefore, the gas pressure inside the tank at 9 degrees Celsius is approximately 43.54 atm.
To calculate the volume at 26 degrees Celsius and 0.95 atmospheres, we can use the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature (in Kelvin)
P2 = Final pressure
V2 = Final volume
T2 = Final temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin. So, for 26 degrees Celsius:
T2 = 26 + 273.15 = 299.15 K
We are given:
P1 = 43.54 atm
V1 = 4.3 L
T1 = 9 degrees Celsius + 273.15 = 282.15 K
P2 = 0.95 atm
Now we can rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
V2 = (43.54 atm * 4.3 L * 299.15 K) / (0.95 atm * 282.15 K)
V2 = 5577.350905 / 266.84325
V2 ≈ 20.91 L
Therefore, the volume occupied by the gas at 26 degrees Celsius and 0.95 atmospheres is approximately 20.91 L.