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Question: The circulation of blood in the human body supplies \({O_2}\) and releases \(C{O_2}\) . The concentr...

The circulation of blood in the human body supplies O2{O_2} and releases CO2C{O_2} . The concentration of O2{O_2} and CO2C{O_2} is variable but on the average, 100 mL100{\text{ }}mL blood contains 0.32 g0.32{\text{ }}g of O2{O_2} and 0.88 g0.88{\text{ }}g CO2C{O_2} . The volume of O2{O_2} and CO2C{O_2} at 1 atm1{\text{ }}atm and body temperature 37OC{37^O}C , assuming 1010 litre blood in human body is:
A.2.545L,2.545L2.545L,2.545L
B.5.09L,2.545L5.09L,2.545L
C.2.545L,5.09L2.545L,5.09L
D.2.545L,7.635L2.545L,7.635L

Explanation

Solution

In the given question firstly we have to define the conditions for the given cases. The two gases talked about are O2{O_2} and CO2C{O_2} . Now we just have to use the ideal gas equation, PV=nRTPV = nRT simply to get the volume of both the gases by putting the other values that are given in the problem.

Complete step by step answer:
First we have to sort out every detail of the given problem statement :
i.The two gases talked about are : O2{O_2} and CO2C{O_2}
ii.The amount of the O2{O_2} in the 100 mL100{\text{ }}mL blood : 0.32 g0.32{\text{ }}g
iii.The amount of the CO2C{O_2} in the 100 mL100{\text{ }}mL blood : 0.88 g0.88{\text{ }}g
iv.Now let suppose the volume of the O2{O_2} and CO2C{O_2} at 1 atm1{\text{ }}atm and body temperature 37oC{37^o}C , assuming 1010 litre blood is xx and yy .
To get the answer we firstly have to transfer the quantity of litre into the millilitres :
The amount of litres are : 1010
So the amount in the milliliters would be :
=10×1000ml =10000ml  = 10 \times 1000ml \\\ = 10000ml \\\
Now we have to use the universal gas equation / ideal gas equation :
PV=nRTPV = nRT
So we have to use that for both the cases:
Case 1: Here the pressure is 1 atm1{\text{ }}atm and the temperature is 310K310K , now we have to use and find the volume :

PV=nRT 1×V=3.232×0.0821×310 V=2.545litre  PV = nRT \\\ 1 \times V = \dfrac{{3.2}}{{32}} \times 0.0821 \times 310 \\\ V = 2.545litre \\\

Case 2: Here the pressure is 1 atm1{\text{ }}atm and the temperature is 310K310K , now we have to use and find the volume :

PV=nRT 1×V=8.844×0.0821×310 V=5.09litre  PV = nRT \\\ 1 \times V = \dfrac{{8.8}}{{44}} \times 0.0821 \times 310 \\\ V = 5.09litre \\\

Therefore the correct option would be option C, 2.545L,5.09L2.545L,5.09L .

Note:
The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions, although it has several limitations. It was first stated by Benoît Paul Émile Clapeyron in 18341834 .