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 one hour the average cockroach running at \(0.08 \mathrm{~km} / \mathrm{hr}\) consumed \(0.8 \mathrm{~mL}\) of \(\overline{\mathrm{O}_{2}}\) at 1 atm pressure and \(24^{\circ} \mathrm{C}\) per gram of insect weight. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in \(1 \mathrm{hr}\) by a \(5.2-\mathrm{g}\) cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a \(1-\) qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than \(20 \%\) of the available \(\mathrm{O}_{2}\) in a \(48-\mathrm{hr}\) period? (Air is 21 mol percent \(\mathrm{O}_{2}\).)

Short answer

In this problem, a 5.2-g cockroach consumes 0.000188 moles of oxygen per hour when running at the given speed. In a 1-quart fruit jar, there are 0.008038 moles of available oxygen. Over a 48-hour period, the cockroach consumes 0.009024 moles of oxygen, which is approximately 112.3% of the available oxygen. Therefore, the cockroach will consume more than 20% of the available oxygen in the 48-hour period.

Step-by-step solution

Calculate the total volume of oxygen consumed per hour by 5.2 g cockroach

According to the problem, the cockroach consumes 0.8 mL of oxygen per gram of weight per hour. Therefore, for a 5.2 g cockroach: Total volume of oxygen consumed per hour = (0.8 mL/g) × 5.2 g Total volume of oxygen consumed per hour = 4.16 mL

Convert the volume of oxygen into moles

We are given that the pressure is 1 atm and the temperature is 24°C, which we need to convert to Kelvin (K): Temperature in Kelvin = 24 + 273.15 = 297.15 K Now we will use the ideal gas law formula, \(PV = nRT\), to find the number of moles of oxygen consumed by the cockroach: \(n = \frac{PV}{RT}\) where; n = moles of oxygen P = pressure in atm = 1 atm V = volume in liters = 4.16 mL (convert mL to L) = 0.00416 L R = ideal gas constant = 0.0821 L atm / (K mol) T = temperature in K = 297.15 K Moles of oxygen consumed by the cockroach in 1 hour: \(n = \frac{(1)(0.00416)}{(0.0821)(297.15)}\) \(n = 0.000188 \ \mathrm{mol}\)

Calculate the moles of oxygen in the jar

We know that the jar is 1-quart in size. Let's convert 1-quart to liters: 1 quart = 0.946352 liters Since air contains 21% oxygen by mol, we need to calculate the moles of air in the jar first, then find the moles of oxygen present in it: Using ideal gas law formula for air in the jar: \(n_{\mathrm{air}} = \frac{PV}{RT}\) Moles of air in the jar: \(n_{\mathrm{air}} = \frac{(1)(0.946352)}{(0.0821)(297.15)}\) \(n_{\mathrm{air}} = 0.03829 \ \mathrm{mol}\) Now, we calculate the moles of oxygen in the jar: Moles of oxygen = 0.21 * Moles of air Moles of oxygen = 0.21 * 0.03829 = 0.008038 \ \mathrm{mol}

Calculate the oxygen consumed in 48 hours

We know that the cockroach consumes 0.000188 moles of oxygen in 1 hour, so we can find the moles of oxygen consumed in 48 hours: Moles of oxygen consumed in 48 hours = 0.000188 * 48 = 0.009024 \ \mathrm{mol}

Check if more than 20% of available oxygen is consumed

Now we will calculate the percentage of oxygen consumed by the cockroach over the 48 hours: Percentage of oxygen consumed = \(\frac{\mathrm{moles\ of\ oxygen\ consumed}}{\mathrm{moles\ of\ oxygen\ in\ the\ jar}} * 100\% \) Percentage of oxygen consumed = \(\frac{0.009024}{0.008038} * 100\% \) Percentage of oxygen consumed = 112.3% Since 112.3% is greater than 20%, yes, the cockroach will consume more than 20% of the available oxygen in the 48-hour period.

Key concepts

Oxygen Consumption

In order to understand oxygen consumption, let's first consider why it's important. All living organisms, including cockroaches, require oxygen to produce energy through respiration. The amount of oxygen consumed can tell us a lot about the metabolic rate and activity levels of an organism.
In this experiment, male cockroaches running on a treadmill consumed a specific amount of oxygen. We know this consumption occurs at a rate of 0.8 mL per gram of cockroach per hour. For a cockroach weighing 5.2 grams, the total volume of oxygen consumed in an hour is multiplied by its weight, amounting to 4.16 mL.
This rate of oxygen intake allows us to explore how active the cockroach is and make predictions about its metabolic needs over time. This simple calculation helps us understand how different factors like weight and activity level affect oxygen consumption.

Mole Calculation

Mole calculation is an essential process used to determine the amount of a substance present in a chemical reaction. It relates the mass of a substance to the number of particles contained using Avogadro's number, which is approximately 6.022 x 10^23 particles per mole.
The ideal gas law is utilized here to find the amount of moles present. The formula, \( PV = nRT \), is rearranged to solve for \( n \) (moles of gas):

• \( P \) is pressure (1 atm)
• \( V \) is volume in liters (converted from 4.16 mL to 0.00416 L)
• \( R \) is the ideal gas constant (0.0821 L atm / K mol)
• \( T \) is temperature in Kelvin (24°C converted to 297.15 K)

Using this method, we determine that the cockroach consumes approximately 0.000188 moles of oxygen per hour. Understanding how to convert volume to moles using conditions of pressure and temperature is vital for tasks like determining metabolic rates or gas production.

Gas Volume Conversion

Gas volume conversion helps us relate different units of volume, such as converting from quarts to liters, which is particularly useful when working with gases. When we assess the oxygen availability in an environment, we often need to convert units to maintain consistency with the Ideal Gas Law formula.
In our scenario, a fruit jar has a volume of 1 quart, which must be converted to liters for easier application with the Ideal Gas Law. Therefore, 1 quart equals 0.946352 liters. Apply the Ideal Gas Law again to find the number of moles of air in the jar, as air is about 21% oxygen by mole percentage. This conversion enables a deeper understanding of how much oxygen is available versus how much is consumed.
Conducting a similar conversion for consumed oxygen over extended periods helps investigate if a cockroach would deplete available oxygen in an enclosed space like a jar. This method is not only applicable here but is widely used in chemical experiments where different gases react or mix under controlled conditions.