Question

A $1.0kg$ball is at rest on the floor in a$2.0m\times 2.0m\times 2.0m$ room of air at $STP$. Air is$80\%$ nitrogen $\left(N_2\right)$and$20\%$ oxygen$\left(O_2\right)$ by volume.
a. What is the thermal energy of the air in the room?
b. What fraction of the thermal energy would have to be conveyed to the ball for it to be spontaneously launched to a height of$1.0m$ ?
c. By how much would the air temperature have to decrease to launch the ball?
d. Your answer to part $c$ is so small as to be unnoticeable, yet this event never happens. Why not?

Short answer

$\left(a\right)$The thermal energy of air in the room is$2\mathrm{MJ};$

$\left(b\right)$The Fraction of thermal energy is$4.9\cdot {10}^{-6};$

$\left(c\right)$The air temperature decreases atlocalid="1648536068001" $1.4\mathrm{mK};$

$\left(d\right)$The molecules in the ball move in all directions equally.

Step-by-step solution

Step: 1  Finding thermal energy: (part a)

The energy can be distributed in a variety of ways.

$E=\frac{5}{2}nRT$

The number $n$of moles from the ideal gas law,

$pV=nRT\Rightarrow n=\frac{pV}{RT}$ $pV=nRT\Rightarrow n=\frac{pV}{RT}$

The energy can be expressed in a variety of ways.

$E=\frac{5}{2}pV$

The degree of freedom are $5$.

$E=\frac{5}{2}\cdot {10}^5\cdot 8=2\cdot {10}^6\mathrm{J}.$

Step: 2  Finding the fraction of thermal energy: (part b)

The work needed will be,

$W=mgh=1\cdot 9.8\cdot 1=9.8\mathrm{J}$

Our required ratio is,

$R=\frac{9.8}{2\cdot {10}^6}=4.9\cdot {10}^{-6}.$

Step: 3 c and d Change in temperature: (part c and d)

We've recently calculated the proportion of energy required for our job to overall energy. The energy is linearly related to the temperature, as we observed in the first equation. This means that the temperature, energy, and energy change will be supplied to the temperature, energy, and energy change, respectively.

$\mathrm{\Delta }T=RT.$

$\mathrm{\Delta }T=4.9\cdot {10}^{-6}\cdot 273=0.00014\mathrm{K}.$

$\left(d\right)$The momentum that the ball receives from the air raises the energy of all molecules. Because these molecules fly in random directions, they have no effect on the ball's movement. The ball would move if they all moved in the same direction.