Question

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In 30 minutes the average cockroach (running at \(0.08 \mathrm{~km} / \mathrm{h})\) consumed \(1.0 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at \(101.33 \mathrm{kPa}\) pressure and \(20^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 day by a 6.3 -g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 2.0-L fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, how much of the available \(\mathrm{O}_{2}\) will the cockroach consume in 1 day? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2} .\) )

Short answer

A 6.3-g cockroach running at 0.08 km/h would consume 0.0122784 moles of O₂ in 1 day. If placed in a 2.0-L fruit jar with a tight lid, assuming the same level of continuous activity, the cockroach would consume approximately 70.83% of the available O₂ in the jar in 1 day.

Step-by-step solution

Determine moles of O₂ consumed per gram of cockroach in 30 minutes

To calculate this, we need to use the given data: Oxygen consumption: 1.0 mL of O₂ per gram of insect mass Pressure: 101.33 kPa Temperature: 20°C (293 K) We can use the ideal gas law formula (PV = nRT) to find the moles of O₂ consumed. Rearranging the formula to find 'n', we get: n = PV / RT To convert mL to L, we must divide by 1000: Volume (V) = 1.0 mL / 1000 = 0.001 L Now, we can substitute the values in the formula: n = (101.33 kPa * 0.001 L) / (8.314 kPa L/mol K * 293 K) n = 0.0000406 mol So, 0.0000406 moles of O₂ were consumed in 30 minutes by 1 gram of cockroach.

Calculate moles of O₂ consumed by a 6.3-g cockroach in 1 day

Now that we know the consumption per gram of cockroach in 30 minutes, we can calculate the consumption for a 6.3-g cockroach in 1 day: 1 day = 24 hours = 1440 minutes Number of 30-minute intervals in 1 day = 1440 minutes / 30 minutes = 48 intervals Since the cockroach's mass is 6.3-g, the moles of O₂ consumed during one 30-minute interval for the entire cockroach is: 0.0000406 mol/g * 6.3 g = 0.0002558 mol Now, we can multiply by the number of 30-minute intervals in 1 day: 0.0002558 mol * 48 = 0.0122784 mol So, the 6.3-g cockroach would consume 0.0122784 moles of O₂ in 1 day while moving at the given speed.

Calculate the moles of available O₂ in the 2.0-L jar

We are given that the air in the jar contains 21% mol/mol of O₂. To find the total moles of air in the jar, we can use the ideal gas law formula and the given temperature and pressure conditions: Volume (V) = 2.0 L Pressure: 101.33 kPa Temperature: 293 K Moles of air (n) = PV / RT n = (101.33 kPa * 2.0 L) / (8.314 kPa L/mol K * 293 K) n = 0.08254 mol Now, we can find the moles of O₂ in the jar: Moles of O₂ = 0.21 * 0.08254 mol Moles of O₂ = 0.0173324 mol

Calculate the percentage of O₂ consumed by the cockroach in 1 day

Now that we have both the moles of O₂ consumed by the cockroach in 1 day (from Step 2) and the moles of available O₂ in the jar (from Step 3), we can find the percentage of O₂ consumed: Percentage of O₂ consumed = (moles of O₂ consumed / moles of available O₂) * 100 Percentage of O₂ consumed = (0.0122784 mol / 0.0173324 mol) * 100 Percentage of O₂ consumed ≈ 70.83% So, the cockroach will consume approximately 70.83% of the available O₂ in the 2.0-L fruit jar in 1 day, assuming the same level of continuous activity as in the research.

Key concepts

Oxygen Consumption in Cockroaches

Understanding oxygen consumption is key in physiology, especially when studying how animals like cockroaches sustain themselves. Oxygen is vital for energy production, and measuring it can show an organism's metabolic rate. In our case, cockroaches were observed while running, measuring the oxygen they used up.

In the given experiment, cockroaches consumed oxygen at a rate of 1.0 mL per gram in 30 minutes. It may seem like a small amount, but for a tiny insect, it's significant. This data allows us to understand how efficient these creatures are at converting oxygen into energy.

Knowing the oxygen consumption is the first step. The next is to link it to how much oxygen is available and calculate consumption over longer periods or larger quantities.

Moles Calculation using the Ideal Gas Law

When you measure gases like oxygen, it's critical to understand moles—a fundamental unit in chemistry. To find out how many moles of oxygen are involved, we use the Ideal Gas Law: \( PV = nRT \) Where: * \( P \) is pressure, * \( V \) is volume, * \( n \) is the number of moles, * \( R \) is the gas constant, * \( T \) is temperature in Kelvin.

In our context, if a cockroach consumes 1 mL of oxygen per gram, we first convert this volume into liters (0.001 L). Using the conditions provided (101.33 kPa and 293 K), we plug into the ideal gas law to derive the moles consumed. The result, 0.0000406 moles of \( O_2 \) per gram every 30 minutes, helps estimate how much oxygen is used over longer periods.

This calculation is crucial because it transforms real-world measurements into chemical data we can work with, allowing deeper understanding of physiological processes over time.

Temperature and Pressure Conditions in Gas Calculations

To calculate the moles of a gas, it’s essential to understand how conditions like temperature and pressure impact gas measurements. Gases behave differently under various circumstances, so we use specific conditions to perform accurate calculations.

In our exercise, the temperature is 20°C, converted to Kelvin (293 K), and pressure is measured at 101.33 kPa. These conditions are foundational in using the Ideal Gas Law. Only with temperature and pressure factored in can we calculate the moles of oxygen accurately.

Consider the conditions like settings on a stage, critical in knowing how the gas behaves. Different temperatures and pressures would yield different results, demonstrating why precision is crucial in measuring and understanding gas consumption.