Question

25) Find the Kinetic Energy of 2 moles of an ideal gas in calaries at 27degree Celsius

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Answer 1

$\boxed{\bold {\green{Answer}}}$

1800 cal

$\boxed{\bold {\green{Explanation}}}$

Given:

• T = 27°c = 300k
• n = 2 moles

To Find:

Kinetic Energy

Solution:

$\boxed{\bold {\red{K.E. = \dfrac{3}{2}nRT}}}$

$\mathsf{K.E. = \dfrac{3}{2}(2)(2)(300)}$

$\mathsf{K.E. = (3)(2)(300)}$

$\mathsf{K.E. = 1800 cal}$

Answer 2

Given:

1. T = 27°c = 300k
2. n = 2 moles

To Find:

Kinetic Energy

Solution:

$\boxed{\bold {\orange{K.E. = \dfrac{3}{2}nRT}}}$

$\tt{K.E. = \dfrac{3}{2}(2)(2)(300)}$

$\tt{K.E. = (3)(2)(300)}$

$\huge\tt{K.E. = 1800 cal}$