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:
41
450
2.5 × 1025
2.7 × 1026
5.4 × 1026
See answer to sam at:
http://www.jiskha.com/display.cgi?id=1245847346
To find the approximate number of air molecules in a 1 m3 volume at room temperature and atmospheric pressure, we can use the ideal gas law equation.
The ideal gas law equation is expressed as:
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 atmospheric pressure to atm. Since the pressure is given in atmospheres, we can directly use it in the equation.
Next, we need to convert the volume from m3 to liters (L) since the units of R are in m3 atm/mol K. There are 1000 L in 1 m3.
Now, we have the pressure (in atm), volume (in L), the gas constant (R), and the temperature (in K). We can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Plugging in the values:
P = 1 atm
V = 1000 L
R = 8.2 × 10^(-5) m^3 atm / mol K (We need to convert it to L atm / mol K. There are 1000 L in 1 m^3)
T = 300 K
n = (1 atm * 1000 L) / (8.2 × 10^(-5) m^3 atm/mol K * 1000 L)
= 1 / (8.2 × 10^(-5) m^3/mol)
≈ 1.22 × 10^(4) mol
To find the number of molecules in moles, we multiply by Avogadro's number (NA):
Number of molecules = n * NA
= 1.22 × 10^(4) mol * 6.02 × 10^(23) molecules/mol
= 7.35 × 10^(27) molecules
Therefore, the approximate number of air molecules in a 1 m3 volume at room temperature and atmospheric pressure is approximately 7.35 × 10^(27). This value closest to this is 5.4 × 10^(26).