calculate r.m.s velocity of he molecule at 300k.also calculate the total k.e(in calories)of the molecules in one litre he at 300k & 1atm?
To calculate the root mean square (r.m.s) velocity of a molecule, you can use the following formula:
v(r.m.s) = √(3kT/m)
where:
v(r.m.s) is the root mean square velocity
k is the Boltzmann constant (approximately 1.38 x 10^-23 J/K)
T is the temperature in Kelvin (300K in this case)
m is the mass of one molecule (assume helium - 6.6464764 x 10^-27 kg)
Using these values, you can calculate the r.m.s velocity:
v(r.m.s) = √(3 * 1.38 x 10^-23 J/K * 300 K / 6.6464764 x 10^-27 kg)
v(r.m.s) ≈ 1930 m/s
To calculate the total kinetic energy (k.e) of the molecules in one liter of helium at 300K and 1 atm, you can use the following formula:
k.e = (1/2) * N * M * (v(r.m.s))^2
where:
k.e is the total kinetic energy
N is the number of molecules (Avogadro's number, approximately 6.022 x 10^23 molecules/mol)
M is the molar mass of helium (4.0026 g/mol)
v(r.m.s) is the r.m.s velocity (calculated earlier)
Converting the units appropriately:
M = 4.0026 g/mol = 4.0026 x 10^-3 kg/mol
1 L = 1000 cm^3 = 10^-3 m^3
Substituting the values into the formula:
k.e = (1/2) * (6.022 x 10^23 molecules/mol) * (4.0026 x 10^-3 kg/mol) * (1930 m/s)^2
k.e ≈ 7.151 x 10^4 J
To convert the energy from Joules to calories, you can use the conversion factor:
1 calorie ≈ 4.184 J
Converting the total kinetic energy:
k.e (in calories) = 7.151 x 10^4 J * (1 calorie / 4.184 J)
k.e (in calories) ≈ 1.71 x 10^4 calories
Therefore, the r.m.s velocity of the molecules at 300K is approximately 1930 m/s, and the total k.e of the molecules in one liter of helium at 300K and 1 atm is approximately 1.71 x 10^4 calories.