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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 0.08km/hr consumed 0.8 mL of O2 per gram of insect weight at 1 atm pressure and 24 ∘C. a)

How many moles of O2 would be consumed in 1 hr by a 5.4?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 O2 in a 48-hr period? (Air is 21 mol percent O2)

Respuesta :

Answer:

Part - A:

[tex]1.8\times 10^{-4}mol[/tex]moles oxygen consumed.

Part - B:

The cockroach needs all the available oxygen in the jar not only 20%.  

Explanation:

Part-A:

From the given,

Mass = 5.4 g

Time = one hour

Rate of consumption = [tex]\frac{0.8molO_{2}}{1g.hr}[/tex]

Volume of [tex]O_{2}[/tex] consumed:

[tex]V= Mass \times time \times rate\,of\,consumption[/tex]

[tex]=5.2g \times 1hr \times \frac{0.8molO_{2}}{1g.hr} = 4.32 ml[/tex]

Therefore, 4.32 ml of oxygen consumed in one hour by a 5.4 of cockroach moving at this speed.

Number of moles of oxygen consumed:

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{1\times 4.32[\frac{1L}{1000mL}]}{0.082\times 297}=0.000177mol=1.8\times 10^{-4}mol[/tex]

Therefore, [tex]1.8\times 10^{-4}mol[/tex]moles oxygen consumed.

Part -B:

From the given,

P = Pressure = 1atm

V = Volume = 4.32 ml = 0.00432 L

R = Gas constant = 0.0821 L.atm/mol.K

T = 24 C = 273 + 24 = 297 K

Consumed moles of oxygen can be calculated by following formula.

[tex]n=\frac{PV}{RT}[/tex]

Substitute the all given values.

[tex]= \frac{1 \times 0.00432}{0.0821 \times 297}= 1.77 \times 10^{-4}mol[/tex]

Volume of available oxygen:

[tex]V= 1qt\,air \times \frac{0.946 L }{1\,qt}\times 0.21 O_{2}= 0.199 L[/tex]

Total moles of oxygen can be calculated by following formula.

[tex]n=\frac{PV}{RT}[/tex]

[tex]= \frac{1 \times 0.199L}{0.0821 \times 297}= 8.16 \times 10^{-3}mol[/tex]

Amount of oxygen used by cockroach:

[tex]= \frac{1.77 \times 10^{-4}mol}{1 hr}\times 48hr = 8.0 \times 10^{-3}mol[/tex]

Hence, The cockroach needs all the available oxygen in the jar not only 20%.  

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