Question

total kinetic energy of 56g of N2 gas at 27 celcius is​

Answer 1

Answer:

Kinetic energy of the ideal gas can be calculated as

Kinetic energy = {3*n*R*T} / 2

where,

n = number of moles

R = gas constant

T = temperature in kelvin

We have above is, n = 5 moles of N

R = 8/314,

and T = 27 + 273 (to convert celsius to kelvin, add 273)=300.

Kinetic energy = {3*5*8.314*300} / 2

= 18,706.05 joules or 18.706 kj

Answer 2

Answer:

Total kinetic energy of 56g of N₂ gas at 27°C is equal to 7.4826KJ.​

Explanation:

Given, the mass of nitrogen gas = 56g

The molecular mass of the nitrogen gas = 28gmol⁻¹

The number of moles of N₂ gas = 56/28= 2 moles

We know the relation of kinetic energy with temperature:

Kinetic energy $=\frac{3}{2}nRT$                                                              ................(1)

where n is number of moles of gas

R is gas constant

T is the temperature.

We have n=2, R= 8.314Jmol⁻¹K⁻¹ and T = 273+27=300K

Put the value of n, r and T in equation (1):

$K.E.=\frac{3}{2} (2mol)(8.314JK^{-1}mol^{-1})(300K)$

$K.E. =7482.6J$

$K.E.=7.4826KJ$

Therefore, the total kinetic energy of nitrogen gas is 7.4826KJ.