Question

A diesel engine works at a high compression ratio to compress air until it reaches a temperature high enough to ignite the diesel fuel. Suppose the compression ratio (ratio of volumes) of a specific diesel engine is 20.0 to 1.00. If air enters a cylinder at 1.00 atm and is compressed adiabatically, the compressed air reaches a pressure of 66.0 atm. Assuming that the air enters the engine at room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) and that the air can be treated as an ideal gas, find the temperature of the compressed air.

Short answer

Answer: The temperature of the compressed air in the diesel engine is approximately 1046.56 K.

Step-by-step solution

Write down the given information

We are given the following information: - Compression ratio: \(V_1 / V_2 = 20 / 1\) - Initial pressure of air: \(P_1 = 1.00\,\text{atm}\) - Initial temperature of air: \(T_1 = 25.0^{\circ} \mathrm{C} = 298.15\,\text{K}\) - Final pressure of compressed air: \(P_2 = 66.0\,\text{atm}\) - Compression process is adiabatic and air can be treated as an ideal gas

Use the adiabatic equation for an ideal gas

Since air can be treated as an ideal gas and the compression process is adiabatic, we can use the following equation: \(T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}\) where \(T_1\) and \(T_2\) are the initial and final temperatures, \(V_1\) and \(V_2\) are the initial and final volumes, and \(\gamma\) (gamma) is the specific heat ratio (also known as adiabatic index) for air, which is approximately 1.4.

Replace the volume ratio by the compression ratio

Since we are given the compression ratio, we know that \(V_1 / V_2 = 20\). This implies that \(V_2 = \frac{1}{20}V_1\). We can now replace the volume terms in the equation: \(T_1 V_1^{\gamma - 1} = T_2 \left(\frac{1}{20}V_1\right)^{\gamma - 1}\) Now, we can divide both sides of the equation by \(V_1^{\gamma - 1}\): \(T_1 = T_2 \left(\frac{1}{20}\right)^{\gamma - 1}\) And we want to find \(T_2\), the final temperature.

Solve for the final temperature T_2

Now, substitute the given values and solve for \(T_2\): \(T_1 = T_2 \left(\frac{1}{20}\right)^{\gamma - 1}\) \(298.15\,\text{K} = T_2 \left(\frac{1}{20}\right)^{1.4 - 1}\) Divide both sides by \(\left(\frac{1}{20}\right)^{0.4}\): \(T_2 = 298.15\,\text{K} \times \left(\frac{1}{20}\right)^{-0.4}\) Calculating the final temperature T_2: \(T_2 \approx 1046.56\,\text{K}\)

Write down the final answer

The temperature of the compressed air in the diesel engine is approximately \(1046.56\,\text{K}\).