Question

A player bounces a basketball on the floor, compressing it to 80.0% of its original volume. The air (assume it is essentially ${\mathbf{N}}_{\mathbf{2}}$gas) inside the ball is originally at 20.0°C and 2.00 atm. The ball’s inside diameter is 23.9 cm. (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal. (b) By how much does the internal energy of the air change between the ball’s original state and its maximum compression?

Short answer

1. The maximum temperature the gas can reach after compression Is $\left(47C°\right)$.
2. $∆\mathrm{U}=333\mathrm{J}$ internal energy of the air change between the ball’s original state and its maximum compression

Step-by-step solution

Adiabatic ideal gas relation.

An adiabatic process in which no heat transfer takes place between a system and its surroundings. Zero heat transfer is an idealization.

${\mathrm{TV}}^{\mathrm{Y}-1}=\mathrm{Constant}$

Where, Y = The ratio of heat capacities$\left[\mathrm{Y}\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{v}}}\right]$

Thus for an initial state$\left({\mathrm{T}}_1{\mathrm{V}}_1\right)$and a final state$\left({\mathrm{T}}_2{\mathrm{V}}_2\right)$

${\mathrm{T}}_1{{\mathrm{V}}_1}^{\mathrm{Y}-1}{\mathrm{T}}_2{{\mathrm{V}}_2}^{\mathrm{Y}-1}$ (adiabatic process, ideal gas)

Calculate the temperature at its maximum compression

1. By compressing the Nitrogen gas inside the basketball due to bounces, the volume decreases from the initial state${\mathrm{V}}_1$to final state${\mathrm{V}}_2$.

Whererepresent 80% of${\mathrm{V}}_1$

So,${\mathrm{V}}_2=0.8{\mathrm{V}}_1$

Consider the gas inside the ball is an ideal gas and compressed adiabatically. As given and we want to calculate the maximum temperature ${\mathrm{T}}_2$. In an adiabatic process, the relation is given by.

${\mathrm{T}}_1{{\mathrm{V}}_1}^{\mathrm{Y}-1}{\mathrm{T}}_2{{\mathrm{V}}_2}^{\mathrm{Y}-1}$……..(1)

$\mathrm{For}{\mathrm{N}}_2\mathrm{gas}\mathrm{Y}=1.40\mathrm{where}\left[\mathrm{Y}=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}\right]$

Substitute all the values in equation 1^st

$$\begin{gathered} {\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{Y}-1} \\ {\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{V}}_1}{0.8{\mathrm{V}}_1}\right)}^{\mathrm{Y}-1} \end{gathered}$$

According to question, Initial temperature ${\mathrm{T}}_1=20\mathrm{C}°\left(293\mathrm{K}\right)$, Initial pressure${\mathrm{P}}_1=2\mathrm{atm}$

$$\begin{gathered} {\mathrm{T}}_2=293\mathrm{K}{\left(\frac{1}{0.8}\right)}^{1.40-1} \\ {\mathrm{T}}_2=320\mathrm{K}\mathrm{In}\mathrm{Celsius}(47\mathrm{C}°) \end{gathered}$$

In Celsius

Therefore, The maximum temperature the gas can reach after compression Is $(47\mathrm{C}°)$.

Calculate the volume of ball

For calculating the internal energy$∆\mathrm{U}$, first we have to calculate the volume of ball which is represent the volume of gas${\mathrm{V}}_1$

As the ball is a sphere and its radius

$\mathrm{r}=\frac{\mathrm{d}}{2}$, According to question diameter of ball d = 23.9cm

So,

$$\begin{gathered} \mathrm{r}=\frac{23.9\mathrm{cm}}{2} \\ \mathrm{r}=11.95\mathrm{cm} \end{gathered}$$

Volume of ball

$$\begin{gathered} \mathrm{V}=\frac{4}{3}{\mathrm{\tau \tau r}}^3 \\ \mathrm{V}=\frac{4}{3}\mathrm{\tau \tau }{\left(11.95\times {10}^{-2}\mathrm{m}\right)}^3 \\ \mathrm{V}=7.15\times {10}^{-3}{\mathrm{m}}^{-3} \end{gathered}$$

The volume of ball is represent the volume of gas ${\mathrm{V}}_1$.

Calculate the internal energy.

To calculate the change in internal energy due to adiabatic process it is given by,

$∆\mathrm{U}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)$

Where ,${\mathrm{C}}_{\mathrm{v}}$= Molar heat capacity and for${\mathbf{N}}_{\mathbf{2}}$gas${\mathrm{C}}_{\mathrm{v}}=20.76\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{k}}$

Also we have to calculate the number of moles . So from ideal gas equation

$\mathrm{n}=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{RT}}_1}$……..(2)

Where R = Universal gas constant$\left[\mathrm{R}=8.314\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{k}}\right]$

According the question${\mathrm{P}}_1$=2atm in Pascal${\mathrm{P}}_1=2\times 1.013\times {10}^5\mathrm{Pa}$

Substitute all the values in equation 2^nd

$\mathrm{n}=\frac{2\times 1.013\times {10}^5\mathrm{Pa}\times 7.15\times {10}^{-3}{\mathrm{m}}^{-3}}{8.314{\displaystyle \frac{\mathrm{J}}{\mathrm{mol}.\mathrm{k}}}\times 293\mathrm{K}}$

n = 0.594mol

Now we calculate the change in internal energy$∆\mathrm{U}$

$$\begin{gathered} ∆\mathrm{U}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \\ ∆\mathrm{U}=0.594\mathrm{mol}\times 20.76\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{k}}\left(320\mathrm{K}-293\mathrm{K}\right) \\ ∆\mathrm{U}=333\mathrm{J} \end{gathered}$$

Therefore,$∆\mathrm{U}=333\mathrm{J}$ internal energy of the air change between the ball’s original state and its maximum compression