What is the volume of dry O2 gas at STP, if it had a volume of 5.00 L at 103 kPa and 300 K?
V2= P1V1T2/P2T1
= 103*5*273/101.325*300
=4.63L
At STP, T = 0 C or 273 K and P = 1 atm or 101.325 kPa.
Assuming the gas ideal, we can use the formula:
P1 * V1 / T1 = P2 * V2 / T2
where
P = pressure
V = volume
T = absolute temperature (in K)
subscripts 1 and 2 = initial and final properties respectively
substituting,
103 * 5 / 300 = 101.325 * (V2) / 273
Now solve for V2. Hope this helps~ `u`
Well, if we're talking about the volume of dry O2 gas, I must say that it's quite "air-ifying"! Now, let's calculate it. So, at 103 kPa and 300 K, we can use the ideal gas law equation: PV = nRT. Given that the initial volume, P, and T are known, we can find the number of moles, n. Dividing the moles by the molar volume at STP (22.4 L/mol), we can then calculate the final volume. But remember, in the world of chemistry, it's always a good idea to "gas up" your calculations! So, after all the calculations, don't be "deflated" when I tell you that the volume of dry O2 gas at STP would be approximately 5.65 L.
To find the volume of dry O2 gas at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kPa)
V = volume (in L)
n = number of moles
R = ideal gas constant (8.31 J/(mol·K))
T = temperature (in Kelvin)
First, let's convert the pressure and temperature to the correct units:
Pressure: 103 kPa = 103000 Pa
Temperature: 300 K
Next, we can rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
Now, let's plug in the values:
n = (103000 Pa) * (5.00 L) / ((8.31 J/(mol·K)) * (300 K))
Calculating this: n = 211.00 mol
Now, we can use the number of moles to find the volume at STP. At STP, the temperature is 273 K and the pressure is 101.325 kPa.
To find the volume, we can use the following relationship:
(V1 / T1) = (V2 / T2)
Where:
V1 = initial volume (5.00 L)
T1 = initial temperature (300 K)
V2 = final volume (unknown)
T2 = final temperature (273 K)
Solving for V2:
V2 = (V1 * T2) / T1
V2 = (5.00 L * 273 K) / 300 K
Calculating this: V2 ≈ 4.55 L
Therefore, the volume of dry O2 gas at STP would be approximately 4.55 L, if it had a volume of 5.00 L at 103 kPa and 300 K.
To determine the volume of dry O2 gas at STP, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in Pascals)
V = volume (in cubic meters)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given values to the SI units (Pascals and Kelvin):
Pressure, P = 103 kPa = 103,000 Pa
Temperature, T = 300 K
We also need to calculate the number of moles of O2 gas. To do this, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Now we can plug in the values:
n = (103,000 Pa) * (5.00 L) / ((8.314 J/(mol·K)) * (300 K))
Next, we need to convert the volume from liters to cubic meters:
5.00 L = 0.005 m³
Now we can substitute all the values into the equation to solve for n:
n = (103,000 Pa) * (0.005 m³) / ((8.314 J/(mol·K)) * (300 K))
Finally, we can calculate the number of moles:
n ≈ 0.0619 mol
Since the gas is at STP (standard temperature and pressure), we know that at STP, 1 mole of any gas occupies approximately 22.4 liters.
Therefore, the volume of dry O2 gas at STP can be calculated by multiplying the number of moles by the molar volume at STP:
Volume at STP = (0.0619 mol) * (22.4 L/mol)
Volume at STP ≈ 1.39 L
So, the volume of dry O2 gas at STP, given the initial conditions, would be approximately 1.39 liters.