Use R = 8.2 × 10–5 m3 atm/mol K and NA = 6.02 × 1023 mol–1. The approximate number of air molecules in a 1 m3 volume at room temperature (300 K) and atomospheric pressure is:
The ideal gas law states:
PV=nRT
where
n=number of moles
R is constant given above (atm/mol K)
so
n=PV/RT
=1 atm. * 1 m3 /(8.2*10-5 m3 atm / mol K * 300 K)
=40.65 moles
Check:
air weighs 1.293 g/l, or 1293 g/m3. Take an average molar mass equal to oxygen (32), we have
1293/32=40.406 moles ≈ 40.65
Since the temperature is 300 K and not at STP, the number of moles is expected to be lower.
The approximate number of molecules is obtained by multiplying the number of moles by the avogadro's constant (Na)
Approximate number of molecules
=40.65 moles * 6.02*10-23 mol-1
=2.45*1025
how can the notion of symmetry be used to estimate the number of marbles in a 1-liter jar
Well, it's time to do some number crunching, but don't worry, I'll make it fun!
First, let's calculate the volume of a single molecule of air. We'll call it "V_molecule."
V_molecule = 1 m^3 / (6.02 x 10^23 molecules)
Now, let's figure out how many molecules are in a 1 m^3 volume at room temperature and atmospheric pressure. We'll call it "N_total."
N_total = (R x N_A x T) / P
Where:
R = 8.2 x 10^-5 m^3 atm/mol K
N_A = 6.02 x 10^23 mol^-1
T = 300 K (room temperature)
P = 1 atm (atmospheric pressure)
N_total = (8.2 x 10^-5 m^3 atm/mol K) x (6.02 x 10^23 mol^-1) x (300 K) / (1 atm)
Now, let's calculate N_total:
N_total = (8.2 x 10^-5) x (6.02 x 10^23) x (300) / (1)
N_total = approximately 1.478 x 10^22 air molecules
So, there you have it! Approximately 1.478 x 10^22 air molecules in a 1 m^3 volume at room temperature and atmospheric pressure. That's a lot of molecules clowning around!
To calculate the approximate number of air molecules in a 1 m^3 volume at room temperature (300 K) and atmospheric pressure, we can use the ideal gas equation:
PV = nRT,
where:
P is the pressure,
V is the volume,
n is the number of moles,
R is the ideal gas constant, and
T is the temperature.
We can rearrange the ideal gas equation to solve for n:
n = PV/RT.
Given:
R = 8.2 × 10^(-5) m^3 atm/mol K,
NA = 6.02 × 10^(23) mol^(-1),
V = 1 m^3, and
T = 300 K.
Step 1: Convert the pressure from atm to Pa
1 atm = 1.01325 × 10^5 Pa
Step 2: Convert the volume from m^3 to liters
1 m^3 = 1000 liters
Step 3: Plug in the values into the equation:
n = (PV)/RT
= ((1.01325 × 10^5) * (1000))/(8.2 × 10^(-5) * 300)
Step 4: Calculate the approximate number of air molecules:
n ≈ ((1.01325 × 10^5) * (1000))/(8.2 × 10^(-5) * 300)
≈ (1.01325 × 10^5 * 10^3)/(8.2 × 10^(-5) * 300)
≈ (1.01325 × 10^8)/(8.2 * 10^(-5) * 300)
≈ (1.01325 × 10^8)/(82 * 10^(-5) * 3)
≈ (1.01325 × 10^8)/(246 * 10^(-5))
≈ (1.01325 × 10^8)/(2.46)
≈ 4.121 × 10^7
Therefore, the approximate number of air molecules in a 1 m^3 volume at room temperature (300 K) and atmospheric pressure is approximately 4.121 × 10^7 molecules.
To find the approximate number of air molecules in a 1 m3 volume at room temperature (300 K) and atmospheric pressure, we can use the ideal gas law.
The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we are given the volume (1 m3), temperature (300 K), pressure (atmospheric pressure), and the value of the ideal gas constant (R = 8.2 × 10–5 m3 atm/mol K).
First, we need to convert the atmospheric pressure into atm. Atmospheric pressure is approximately 1.01325 × 105 Pa. To convert Pa to atm, we divide by 1 atm = 101325 Pa.
Pressure (P) = 1.01325 × 105 Pa / 101325 Pa/atm = 1 atm
Now we can rearrange the ideal gas law to solve for the number of moles (n):
n = PV / RT
Substituting the given values:
n = (1 atm) * (1 m3) / (8.2 × 10–5 m3 atm/mol K * 300 K)
Simplifying:
n = 3.66 × 103 mol
Finally, using Avogadro's number (NA = 6.02 × 1023 mol–1), we can find the approximate number of air molecules:
Number of air molecules = n * NA
Number of air molecules = (3.66 × 103 mol) * (6.02 × 1023 mol–1) = 2.2 × 1027 molecules
Therefore, the approximate number of air molecules in a 1 m3 volume at room temperature and atmospheric pressure is approximately 2.2 × 1027 molecules.