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Question

The composition of air inhaled by a human is 21% O2 by volume and 0.03% CO2 and that of exhaled air is 16% O2 and 4.03% CO2. Assuming a typical volume of 6200 L day1, moles of O2 used and CO2 generated by the body at 37 C and 1.00 bar are:
(Assume ideal behaviour)

A
Moles of O2 usedMoles of CO2 generated9.649.64
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B
Moles of O2 usedMoles of CO2 generated12.0412.04
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C
Moles of O2 usedMoles of CO2 generated12.049.64
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D
Moles of O2 usedMoles of CO2 generated9.6412.04
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Solution

The correct option is C Moles of O2 usedMoles of CO2 generated12.049.64
We know,
PV=nRT
n=PVRT
Given,
P=1 bar, T=37+273=301 K, R=0.083 bar L mol1K1

Moles of O2 used=Moles of [O2(inhaled)O2(exhaled)]
Moles of O2 used=1×62000.083×310[2110016100]=12.04 mol

Moles of CO2 generated=Moles of [CO2(exhaled)CO2(inhaled)]
Moles of CO2 generated=1×62000.083×310[4.030.03100]=9.64 mol

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