Question: An automobile tyre has a volume of $\mathbf{1}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{\text{m}}}^{\mathbf{3}}$and contains air at a gauge pressure (pressure above atmospheric pressure) of 165 kPawhen the temperature is$\mathbf{0}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}\mathit{C}$.What is the gauge pressure of the air in the tires when the temperature rises to$\mathbf{27}.{\mathbf{0}}^{\mathbf{0}}\mathit{C}$and its volume increases to$\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{\text{m}}}^{\mathbf{3}}$? Assume atmospheric pressure is$\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}\mathbf{.}$

1. Standard air pressure =$101\text{kPa}$
2. Initial (absolute) pressure,$p_i=165+101=266\text{kPa}$
3. Initial volume,$V_i=1.64\times {10}^{-2}{\text{m}}^3$
4. Final volume,$V_f=1.67\times {10}^{-2}{\text{m}}^3$
5. Initial temperature,$T_i=0°\text{C}=273\text{K}$
6. Final temperature,$T_f=27°\text{C}=300\text{K}$

Consider the gas is ideal, use the ideal gas law to relate its parameters, both in the initial state iand in the final state f. In this problem, number of moles, n_i = n_fis the same. So, they cancel out accordingly.

Formula is as follow:

$p_f=\frac{p_i{T}_f{V}_i}{T_i{V}_f}$

Here, P is pressure, V is volume and T is temperature.

To find the final gauge pressure,

$p_f=\frac{p_i{T}_f{V}_i}{T_i{V}_f}$

Substitute the values and solve as:

$$\begin{gathered} p_f=\frac{\left(266\right)\left(1.64\times {10}^{-2}\right)\left(300\right)}{\left(1.67\times {10}^{-2}\right)\left(273\right)} \\ {p}_f=287\text{kPa} \end{gathered}$$

Consider the absolute final pressure. But, as it is expressed as gauge pressure, subtract.$101\text{kPa}$

Therefore, gauge pressure is,

$287\text{kPa}-101\text{kPa}=186\text{kPa}$

Hence,the gauge pressure of the air in the tires is 186 kPa.