Question

The engine of a Ferrari F355 F1 sports car takes in air at $\mathbf{20}\mathbf{°}\mathit{C}$ and 1 atm and compresses it adiabatically to 0.09 times the original volume. The air may be treated as an ideal gas with $\mathit{\gamma }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}$ . (a) Draw a pV-diagram for this process. (b) Find the final temperature and pressure.

Short answer

a.

b. The final temperature ans pressure is $T_2=768K$and $P_2=29atm$respectively.

Step-by-step solution

Draw the pV diagram of the process

For the adiabatic compression process, the volume decreases while the pressure increases. The graph will be as follows

Calculate the final temperature

For an adiabatic process, the relation between temperature and volume is${\mathit{T}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}^{\mathbf{\gamma }\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{T}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}^{\mathbf{\gamma }\mathbf{-}\mathbf{1}}$and between pressure and volume is ${\mathit{P}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}^{\mathbf{\gamma }}\mathbf{=}{\mathit{P}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}^{\mathbf{\gamma }}$ .

Here, $v_1$and $v_2$is the initial and final volume respectively. Next $T_1$ and $T_2$ is the initial and final temperature respectively. And $P_1$ and $P_2$ is the initial and final pressure respectively.

Consider the given data as below.

$V_2=0.09V_1$.

Define the final temperature as below.

$$\begin{gathered} T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1} \\ =293{\left(\frac{V_1}{0.09V_1}\right)}^{14-1} \\ =768K \end{gathered}$$

Calculate the final pressure

Define the final pressure as below.

$$\begin{gathered} P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma } \\ =1\times {\left(\frac{V_1}{0.09V_1}\right)}^{14} \\ =29atm \end{gathered}$$

Result:

The final temperature and pressure is$T_2=768K$ and $P_2=29atm$ respectively.