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 compression ratio.

Step-by-Step Solution

Answer

The final pressure in the cylinder is approximately 3.88 atm.

Step 1: Identify the Variables

Let's identify what we know in the problem. The initial volume \( V_1 = 400 \, \text{cm}^3 \), the final volume \( V_2 = 50.0 \, \text{cm}^3 \), the initial pressure \( P_1 = 1.00 \, \text{atm} \), the initial temperature \( T_1 = 15^{\circ} \text{C} = 288.15 \, \text{K} \), and the final temperature \( T_2 = 77^{\circ} \text{C} = 350.15 \, \text{K} \). We need to find the final pressure \( P_2 \).

Step 2: Convert Celsius to Kelvin

We convert the temperatures from Celsius to Kelvin. The formula is \( T(\text{K}) = T(\text{°C}) + 273.15 \). So, \( T_1 = 15 + 273.15 = 288.15 \, \text{K} \) and \( T_2 = 77 + 273.15 = 350.15 \, \text{K} \).

Step 3: Apply the Combined Gas Law

The Combined Gas Law is given by \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). This relates the initial and final states of pressure, volume, and temperature.

Step 4: Rearrange the Equation

Rearrange the formula to solve for the final pressure: \( P_2 = P_1 \frac{V_1}{V_2} \frac{T_2}{T_1} \).

Step 5: Substitute Known Values and Calculate

Substitute the known values into the equation: \[P_2 = 1.00 \, \text{atm} \times \frac{400 \, \text{cm}^3}{50.0 \, \text{cm}^3} \times \frac{350.15 \, \text{K}}{288.15 \, \text{K}}\] Calculating this gives \(P_2 \approx 3.88 \, \text{atm}\).

Key Concepts

Compression Ratio

In the context of an automotive engine, the "compression ratio" is an important concept. It represents the ratio between the initial and final volumes of the engine cylinder during the compression stroke. A typical compression ratio can be found by dividing the starting volume by the ending volume. In our exercise, the initial volume is 400 cm³ and the final volume is 50 cm³. By dividing these, we find the compression ratio to be 8:1.
This ratio indicates how much the engine compresses the air-fuel mixture. A higher compression ratio usually means more power output from the engine, as it indicates the air is being compressed into a much smaller space. This compression helps in improving engine efficiency and performance.

• High compression ratios can lead to better fuel efficiency.
• They can also increase engine power.
• However, higher compression requires better fuel quality to prevent knocking.

Pressure and Volume Relationship

The relationship between pressure and volume is governed by Boyle's Law, which is a part of the larger context of the combined gas laws. Boyle's Law states that the pressure of a gas is inversely proportional to its volume, provided the temperature remains constant. As the volume of a gas decreases, its pressure increases, assuming the amount of gas and temperature stay the same.
In our exercise, the gas within the engine's cylinder is compressed from 400 cm³ to 50 cm³. This dramatic reduction in volume results in an increase in pressure. The compressed gas exerts more pressure on the cylinder walls, which is essential for the engine's power stroke.

• Decreasing volume increases pressure if temperature is constant.
• Increasing volume decreases pressure if temperature is constant.
• This principle helps understand engine efficiency dynamics.

Temperature Conversion

When dealing with gas laws, it's important to use the absolute temperature scale, which is Kelvin. Celsius temperatures must be converted to Kelvin to apply the gas laws accurately. The relationship is simple: add 273.15 to the Celsius temperature to get Kelvin.
In our scenario, the initial and final temperatures were given in Celsius. We converted 15°C to 288.15 K and 77°C to 350.15 K. Using Kelvin ensures that calculations using gas laws, like the Combined Gas Law, remain consistent and correct.

• Celsius to Kelvin conversion: add 273.15.
• Kelvin is used in gas calculations for accuracy.
• Always perform conversion before calculations.

Ideal Gas Law

The Ideal Gas Law combines several fundamental measures of a gas into one equation: pressure, volume, temperature, and amount of gas. Though we are focusing on the Combined Gas Law in this exercise, the principles are related.
For our problem, the Combined Gas Law is applicable since we are not altering the quantity of gas, but we are changing pressure, volume, and temperature. The formula for the Combined Gas Law is \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]This states that the ratio of the product of pressure and volume to the temperature is constant.

• Relates pressure, volume, and temperature changes.
• Assumes constant number of gas moles.
• Derived from the Ideal Gas Law for dynamic conditions.

Understanding these concepts is essential for solving real-world physics problems like the one in this engine scenario, where multiple factors interact based on these fundamental gas laws.