Question

Calculate the kinetic energy of 2 moles of an ideal gas at 24degree celcius

Answer 1

Answer:

74.042 J

Explanation:

Given :

Numbers of moles = 2 moles

Temperature = 24 C

In Kelvin , T = 273 + 24 = 297 K

Now , we have to calculate K.E.

We have formula :

K.E. ( average ) = 3 / 2 n R T

Values of R = 0.0831

Now putting values in formula :

K.E. = 3 / 2 × 2 × 0.0831 × 297 J

K.E. = 3 × 0.0831 × 297 J

K.E. = 74.042 J

Hence we get answer.

Answer 2

$\huge\underline\purple{\sf Answer:-}$

߷$\large{\boxed{\sf KE=7.4×{10}^{3}J}}$

$\huge\underline\purple{\sf Solution:-}$

TO CALCULATE KINETIC ENERGY WE HAVE FORMULA i.e :-

߷$\large{\boxed{\sf KE={\frac{3}{2}}nRT}}$

Here ,

n = moles

R = Gas Constant

T = Temperature

KE = Kinetic Energy

߷Given :-

Mole (n) = 2 moles

Temperature (T)=24°C

Convert it into Kelvin

Add 273 to the given temperature to convert it into Kelvin

24+273= 297 K

R = 8.314 J/mol.K

KE = ?

On putting value :-

$\large{\sf KE ={\frac{3}{2}}×2×8.314×297}$

$\large\implies{\sf 7407.77}$

߷$\large\red{\boxed{\sf KE=7.4×{10}^{3}J}}$