Question

How many nitrogen molecules are in a 8.86 L container of nitrogen gas at a pressure of 112.1 kPa and a temperature of 30.8 oC?

Answer 1

We will use this law PV = nRT
we have to convert pressure from kPa to atm
1 atm = 101.325 kPa
? atm = 112.1 kPa
P = 1.11 atm
T = 30.8 °C + 273.15 = 303.95 K
V = 8.86 L
R = 0.088205 atm. L / mol . Kelvin
n =
`(PV)/(RT) = (1.11 * 8.86)/(0.08205 * 303.95)` = 0.39 mol
number of N₂ molecules = number of moles * Avogadro's number
= 0.39 * (6.022 * 10²³) = 2.37 x 10²³ molecules

Answer 2

we can first find the number of moles of nitrogen gas in the container using th ideal gas law equation
PV = nRT
where P - pressure - 112 100 Pa
V - volume - 8.86 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 30.8 °C + 273.15 = 303.95 K
substituting these values in the equation
112 100 Pa x 8.86 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 303.95 K
n = 0.393 mol
number of N₂ moles are 0.393 mol

In 1 mol of N₂ there are 6.022 x 10²³ molecules of N₂
therefore in 0.393 mol - 6.022 x 10²³ mol⁻¹ x 0.393 mol = 2.37 x 10²³
there are 2.37 x 10²³ molecules of N₂