Problem 13 - Chemistry and Chemical Reactivity

Question

One of the cylinders of an automobile engine has a volume of \(400 . \mathrm{cm}^{3} .\) The engine takes in air at a pressure of 1.00 atm and a temperature of \(15^{\circ} \mathrm{C}\) and compresses the air to a volume of \(50.0 \mathrm{cm}^{3}\) at \(77^{\circ} \mathrm{C}\). What is the final pressure of the gas in the cylinder? (The ratio of before and after volumes-in this case, 400: 50 or \(8: 1-\) is called the compresion ratio.)

Step-by-Step Solution

Answer

The final pressure is approximately 9.72 atm.

Step 1: Analyze the Given Information

You're given initial and final conditions of air in an engine cylinder: initial volume \(V_1 = 400 \, \text{cm}^3\), initial pressure \(P_1 = 1.00 \,\text{atm}\), and initial temperature \(T_1 = 15^{\circ} \text{C}\) which is \( 288 \text{K} \) after converting to Kelvin. The air is compressed to a final volume \(V_2 = 50.0 \, \text{cm}^3\) and a final temperature \(T_2 = 77^{\circ} \text{C}\) which is \(350 \text{K}\). You need to find the final pressure \(P_2\).

Step 2: Convert Temperatures to Kelvin

Before using the ideal gas law, ensure all temperatures are in Kelvin. Convert \(T_1 = 15^{\circ} \text{C} \) to Kelvin: \(T_1 = 15 + 273 = 288 \, \text{K}\). Similarly, convert \(T_2 = 77^{\circ} \text{C} \) to Kelvin: \(T_2 = 77 + 273 = 350 \, \text{K}\).

Step 3: Use Ideal Gas Law to Relate States

The ideal gas law states \(PV = nRT\). Since the problem deals with the same amount of gas, you can use the combined gas law:\[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \]Rearrange to solve for \(P_2\):\[ P_2 = \frac{P_1 \cdot V_1}{T_1} \cdot \frac{T_2}{V_2} \]

Step 4: Substitute Known Values into Equation

Substitute \(P_1 = 1.00 \,\text{atm}, V_1 = 400 \, \text{cm}^3, T_1 = 288 \, \text{K}, V_2 = 50.0 \, \text{cm}^3, T_2 = 350 \, \text{K}\) into the equation:\[P_2 = \left(\frac{1.00 \, \text{atm} \times 400 \, \text{cm}^3}{288 \, \text{K}}\right) \times \frac{350 \, \text{K}}{50.0 \, \text{cm}^3}\]

Step 5: Calculate the Result

Perform the calculation:\[ P_2 = \left( \frac{400}{288} \right) \times \left( \frac{350}{50.0} \right) \times 1.00 \,\text{atm} \]\[ P_2 = \left(1.3889\right) \times \left(7\right) \approx 9.72 \, \text{atm} \]

Step 6: State the Final Answer

The final pressure of the gas in the cylinder after compression is approximately \(9.72 \, \text{atm}.\)

Key Concepts

Compression Ratio

In the context of an engine, the compression ratio is a critical concept. It is defined as the ratio between the initial volume and the final volume to which the gas is compressed in the cylinder. For instance, in our example from the engine, the initial volume is 400 cm³, and the final volume after compression is 50 cm³. This gives us a compression ratio of \[CR = \frac{V_{initial}}{V_{final}} = \frac{400\, \text{cm}^3}{50\, \text{cm}^3} = 8:1\]The significance of the compression ratio lies in its influence on the engine’s efficiency and performance. Generally, a higher compression ratio is associated with better thermal efficiency and more power output from an engine of a given size. This is because higher compression allows the engine to extract more mechanical energy from a given amount of fuel-air mixture.

• A high compression ratio means more squeezing of the air-fuel mixture, which can lead to increased power.
• This squeezing improves the efficiency since more of the energy from the fuel is converted into useful work.

A practical understanding of the compression ratio is essential for mechanical engineers who aim to design engines with optimal performance.

Thermodynamic Processes

Thermodynamic processes in engines often involve transformations like expansion or compression of gases, which relate to changes in pressure, volume, and temperature. In this problem, we deal with an isentropic process — a common assumption for idealized engine cycles. In such a situation, an ideal gas law can sometimes simplify calculations.
The ideal gas law is expressed as \[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature. For any given confined mass of gas, variations in \( P \), \( V \), and \( T \) are interconnected.
In an engine, during the compression stroke, air is compressed to a smaller volume causing temperature and pressure to rise. The combined gas law, derived under the assumption of a constant quantity of gas, allows us to relate initial and final states of the gas:

• This combined law simplifies to \( \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \), assuming no gas loss.
• Rearranging gives the capability to solve for unknowns when given parts of the state properties.

Understanding these processes is key to determining how changes in one property affect the others, ensuring engines are designed for both power and efficiency.

Cylinder Volume

The concept of cylinder volume is fundamental when working with engines and is central to understanding how engines operate. It's the space the piston travels within the cylinder, determining the engine's displacement capacity. The displacement, in turn, affects the engine's power and efficiency characteristics.
In mathematical terms, for simple geometry like a cylindrical piston, it's calculated as \[ V = \pi r^2h \]where \( r \) is the radius, and \( h \) is the height or stroke length of the cylinder. However, in problem-solving scenarios like the original exercise, you're often directly given the starting and ending volumes, making the ratio calculations straightforward.
Considerations about cylinder volume include:

• Larger cylinder volume typically means more fuel-air mixture, potentially more power.
• Reduction in volume, as air is compressed, leads to increased pressure and temperature changes.

Practical applications involve analyzing how these volumes relate to other characteristics like compression ratios and thermal changes, thereby optimizing engine cycles for specific performance goals.