Question

Calculate the molecular kinetic energy of one gram of helium at 127 degree celsius

Answer 1

HLO MATE HERE IS UR ANSWER.

Molecular kinetic energy of 1 gran of helium = 3Rt/2M

Here R = 8.32J/mol × k, T = 273 + 127 =400k and M = 4

therefore K.E = 3/2 × 8.31 × 400 /4

= 3 × 831/2

= 1246.5J. = 1246J.

Answer 2

Given: One gram of helium at 127 degrees Celsius.

To find: We have to find the molecular kinetic energy.

Solution:

The molecular kinetic energy is given as-

E=3/2nRT

Where n is the number of moles.

The molecular mass of He is 4 grams.

Thus n=1/4

n=0.25

R is the Rydberg constant which is equal to 8.314J per kelvin per mole.

T is the temperature.

T=127+273

T=400K.

Putting these values in the above equation we get-

E=3/2×8.314×400×0.25

E=1247.1

The molecular kinetic energy is equal to 1247.1K.