# 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.

• 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