Question

calculate the kinetic energy of 2 moles of nitrogen at 27 degree's Celsius

Answer 1

Heya,

here is your answer

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Answer 2

The kinetic energy = 7482.6 J

Given :

2 moles of nitrogen at 27°C

To find :

The kinetic energy

Solution :

Step 1 of 2 :

Write down the given data

By the given

n = 2 moles

$\displaystyle \sf{ R = 8.314 \: J{K }^{ - 1} {mol}^{ - 1} }$

T = 27°C = (27 + 273)K = 300K

Step 2 of 2 :

Calculate the kinetic energy

The kinetic energy

$\displaystyle \sf{ = \frac{3}{2} nRT }$

$\displaystyle \sf{ = \frac{3}{2} \times 2 \times 8.314 \times 300 \: \: J}$

$\displaystyle \sf{ = 7482.6 \: J}$

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