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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A bicycle tire has a pressure of ${\mathbf{7}}{\mathbf{.}}{\mathbf{00}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{}}{{\mathbf{\text{N/m}}}}^{{\mathbf{2}}}$ at a temperature of $\mathbf{18}\mathbf{.}\mathbf{0}{\mathbf{}}^{\mathbf{0}}\mathbf{C}$and contains $\mathbf{\text{2.00 L}}$ of gas. What will its pressure be if you let out an amount of air that has a volume of ${\mathbf{\text{100 cm}}}^{\mathbf{3}}$ at atmospheric pressure? Assume tire temperature and volume remain constant.

The pressure of the gas becomes$6.93×{10}^{5}\mathrm{N}/\mathrm{m}$.

See the step by step solution

## Step 1: State the ideal gas law.

According to the ideal gas law,$\mathbf{\text{PV=NkT}}$

Here, is the pressure of the gas, is the volume occupied by the gas, is the number of molecules in the gas, is the Boltzmann constant, and is the absolute temperature.

## Step 2: Calculate the number of molecules present in the gas initially.

Let be the initial pressure, be the number of molecules present initially, be the final pressure, and be the number of molecules at final pressure. Since volume and temperature are constant, the initial and final temperatures can be denoted as Likewise, initial and final volume can be represented as .

${\mathrm{P}}_{0}=7.00×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(\mathrm{K}\right)=18.{0}^{0}\mathrm{C}+273.15\phantom{\rule{0ex}{0ex}}\mathrm{T}=291.15\mathrm{K}\phantom{\rule{0ex}{0ex}}\mathrm{k}=1.38×{10}^{-23}\mathrm{J}/\mathrm{K}$

role="math" localid="1654860621975" $\mathrm{V}=\frac{2.00\mathrm{L}}{1000\mathrm{L}/{\mathrm{m}}^{3}}\phantom{\rule{0ex}{0ex}}=2×{10}^{-3}{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}$

Substitute the values in the equation

$7.00×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}×2.00×{10}^{-3}{\mathrm{m}}^{3}={\mathrm{N}}_{\mathrm{i}}×1.38×{10}^{-23}\mathrm{J}/\mathrm{K}×291.15\mathrm{K}\phantom{\rule{0ex}{0ex}}{\mathrm{N}}_{\mathrm{i}}=\frac{7.00×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}×2.00×{10}^{-3}{\mathrm{m}}^{3}}{1.38×{10}^{-23}\mathrm{J}/\mathrm{K}×291.15\mathrm{K}}\phantom{\rule{0ex}{0ex}}{\mathrm{N}}_{\mathrm{i}}=3.48×{10}^{23}$

Therefore, the number of molecules present in the gas initially is $3.48×{10}^{23}$.

## Step 3: Calculate the number of molecules removed from the tire.

Let the volume of the air removed from the air be ${\mathrm{V}}_{1}$ and its pressure will be equal to the atmospheric pressure. It can be denoted as ${\mathrm{P}}_{1}$. The number of molecules removed from the tire can be denoted as ${\mathrm{N}}_{1}$.

${\mathrm{P}}_{1}=1.013×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{1}=\frac{100{\mathrm{cm}}^{3}}{{10}^{6}{\mathrm{cm}}^{3}/{\mathrm{m}}^{3}}\phantom{\rule{0ex}{0ex}}=100×{10}^{-6}{\mathrm{m}}^{3}$

Substitute these values in the ideal gas equation.

$1.013×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}×100×{10}^{-6}{\mathrm{m}}^{3}={\mathrm{N}}_{1}\mathrm{x}1.38×{10}^{-23}\mathrm{J}/\mathrm{K}×291.15\mathrm{K}\phantom{\rule{0ex}{0ex}}{\mathrm{N}}_{\mathrm{i}}=\frac{1.013×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}×100×{10}^{-6}{\mathrm{m}}^{3}}{1.38×{10}^{-23}\mathrm{J}/\mathrm{K}×291.15\mathrm{K}}\phantom{\rule{0ex}{0ex}}{\mathrm{N}}_{\mathrm{i}}=2.52×{10}^{21}$

Therefore, the number of molecules removed from the tire is $2.52×{10}^{21}$ .

## Step 4: Calculate the final pressure.

Since volume and temperature are constant, the unknown quantity that remains to calculate the pressure is the number of molecules. The number of molecules present at this stage is the difference between the number of molecules present initially and the number of molecules removed from the tire.

${\mathrm{N}}_{\mathrm{f}}={\mathrm{N}}_{0}-{\mathrm{N}}_{\mathrm{i}}\phantom{\rule{0ex}{0ex}}=3.48×{10}^{23}-2.52×{10}^{21}\phantom{\rule{0ex}{0ex}}=3.45×{10}^{23}$

Substitute these values in the ideal gas equation.

${\mathrm{P}}_{\mathrm{f}}×2×{10}^{-3}{\mathrm{m}}^{3}=3.45×{10}^{23}×1.38×{10}^{-23}\mathrm{J}/\mathrm{K}×291.15\mathrm{K}\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{\mathrm{f}}=\frac{3.45×{10}^{23}×1.38×{10}^{-23}\mathrm{J}/\mathrm{K}×291.15\mathrm{K}}{2×{10}^{-3}{\mathrm{m}}^{3}}\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{\mathrm{f}}=6.93×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}$

Therefore, the final pressure is $6.93×{10}^{5}\mathrm{N}/{\mathrm{m}}^{2}$.