Air in a diesel engine cylinder is quickly compressed from an initial temperature of \(20.0^{\circ} \mathrm{C}\), an initial pressure of \(1.00 \mathrm{~atm}\), and an initial volume of \(600 . \mathrm{cm}^{3}\) to a final volume of \(45.0 \mathrm{~cm}^{3}\). Assuming the air to be an ideal diatomic gas, find the final temperature and pressure.

We are given the following information: Initial temperature: \(T1 = 20.0^{\circ}\mathrm{C} = (20.0 + 273.15)\mathrm{K} = 293.15\mathrm{K}\) Initial pressure: \(P1 = 1.00\mathrm{atm} \approx 1.01 \times 10^5 \mathrm{Pa}\) Initial volume: \(V1 = 600\mathrm{cm}^3 = 6.0 \times 10^{-4} \mathrm{m^3}\) Final volume: \(V2 = 45.0\mathrm{cm}^3 = 4.5 \times 10^{-5} \mathrm{m^3}\)

For an ideal diatomic gas undergoing an adiabatic process, the following relationship applies: \(\left(\frac{V_2}{V_1}\right)^{\gamma - 1} = \frac{T_2}{T_1}\) where \(\gamma = C_p/C_v = 7/5\) for a diatomic gas, \(T_1\) and \(V_1\) are the initial temperature and volume, and \(T_2\) and \(V_2\) are the final temperature and volume.

Using the adiabatic condition, we can find the final temperature \(T_2\): \(\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1}\) Since we are looking for \(T_2\), we can rearrange the equation to get: \(T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1}\) Now, plug in the given values: \(T_2 = 293.15\ \mathrm{K} \times \left(\frac{6.0 \times 10^{-4}\ \mathrm{m^3}}{4.5 \times 10^{-5}\ \mathrm{m^3}}\right)^{7/5 - 1}\) \(T_2 \approx 902.27\ \mathrm{K}\)

Now that we have the final temperature, we can use the ideal gas law to find the final pressure \(P_2\). Since both the initial and the final states are the same gas, we can assume that the number of moles \(n\) and the gas constant \(R\) are the same for both. So, we can write the relationship between the initial and final states as: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We are looking for \(P_2\), so we can rearrange the equation to get: \(P_2 = P_1V_1 \frac{T_2}{T_1V_2}\) Now, plug in the given values: \(P_2 = 1.01 \times 10^5\mathrm{Pa}\times 6.0 \times 10^{-4}\mathrm{m^3}\times\frac{902.27\mathrm{K}}{293.15\mathrm{K}\times 4.5 \times 10^{-5}\mathrm{m^3}}\) \(P_2 \approx 3.61\times 10^6\ \mathrm{Pa}\) The final temperature and pressure after compression are approximately \(902.27\ K\) and \(3.61\times 10^6\ \mathrm{Pa}\), respectively.