What is the total kinetic energy of 1.70 moles of nitrogen gas at 30^oC?

Question:

What is the total kinetic energy of 1.70 moles of nitrogen gas at {eq}30^oC? {/eq}

Kinetic Energy:

The Kinetic energy refers to the energy which is generated by the motion of the object. However, it can be determined by multiplying the mass of the object with its velocity and then divide by 2. Thus, the resulting number is the energy generated by the object.

Answer and Explanation:

  • The Kinetic energy = Kinetic Energy = 1545.3 cal

Calculation:

Kinetic Energy = {eq}\frac{3}{2}nRT {/eq}

Number of moles (n) = 1.70 moles

Temperature (T) = 30 + 273 = 303K

{eq}R = 2 Cal \ mol^{-1} K^{-1} {/eq}

Kinetic Energy ={eq}\frac{3}{2}nRT {/eq}

Kinetic Energy ={eq}\frac{3}{2} \times 1.70 \times 2 \times 303 {/eq}

Kinetic Energy = 1545.3 cal


Learn more about this topic:

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Kinetic Energy: Examples & Definition

from General Studies Science: Help & Review

Chapter 4 / Lesson 14
60K

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