Question

In an experiment, 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 km/hr consumed 0.76 mL of O₂ at 1 atm pressure and 32 °C per gram of insect mass. How many moles of O₂ would be consumed in 1 hr by a 5.4 g cockroach moving at this speed?

Answer 1

`1.64* 10^(-4)` moles of O₂ would be consumed in 1 hr by a 5.4 g cockroach moving at this speed.

Step-by-step explanation:

Volume of oxygen gas consumed by the cockroach in an hour = V

V = 0.76 mL = 0.00076 L

Pressure of the oxygen gas = P = 1 atm

Temperature of the gas = T = 32°C = 305.15 K

Moles of oxygen gas consumed by cockroach in an hour = n

`PV=nRT` (Using ideal gas law)

`n=(1 atm* 0.00076 L)/(0.0821 atm L/molK* 305.15 K)`

`n = 3.0336* 10^(-5) mol`

Moles of oxygen gas consumed by 1 gram of mass of an insect = n

Total moles of oxygen gas consumed by 5.4 grams of an insect is:

`n* 5.4=3.0336* 10^(-5) mol* 5.4 =1.64* 10^(-4) mol`

`1.64* 10^(-4)` moles of O₂ would be consumed in 1 hr by a 5.4 g cockroach moving at this speed.