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Jagdish Feroz Patel
Asked: 2022-11-03T01:35:23+05:30 In:

The circulation of blood in human body supplies `O_(2)` and releases `CO_(2)` The concentration of `O_(2)` and `CO_(2)` is variable but on the average `100mL` blood contains `0.02 g` of `O_(2)` and `0.08 g` of `CO_(2)` Calcultate the volume of `O_(2)` and `CO_(2)` at 1 atm and body temperature `37^(@)C` assuming 10 litre blood in human body .
A. `2 L, 4 L`
B. `1.5L, 4.5 L`
C. `1.59 L, 4.62 L`
D. `3.82 L`, `4.62 L`

The circulation of blood in human body supplies `O_(2)` and releases `CO_(2)` The concentration of `O_(2)` and `CO_(2)` is variable but on the average `100mL` blood contains `0.02 g` of `O_(2)` and `0.08 g` of `CO_(2)` Calcultate the volume of `O_(2)` and `CO_(2)` at 1 atm and body temperature `37^(@)C` assuming 10 litre blood in human body .
A. `2 L, 4 L`
B. `1.5L, 4.5 L`
C. `1.59 L, 4.62 L`
D. `3.82 L`, `4.62 L`
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  1. Correct Answer – C
    `:. 100 ml` blood has `0.02gO_(2)` and `8gCO_(2)`
    `:. 10,000 ml` blood has `2gO_(2)` and `8gCO_(2)`
    Using `PV= nRT`
    for `O_(2), 1xxV_(O_(2))=(2)/(32)xx0.0821xx310`
    `implies V_(o_(2))= 1.59 Litre`
    for `CO_(2), 1xxV_(CO_(2))=(8)/(44)xx0.0821xx310`
    `implies V_(O_(2))= 4.62 Litre`

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Urmila Alex Mann
Asked: 2022-10-29T01:01:57+05:30 In:

The circulation of blood in human body supplies `O_(2)` and releases `CO_(2)`. The concentration of `O_(2)` and `CO_(2)` is variable but on the average, 100 mL blood contains 0.02 g of `O_(2)` and 0.08 g of `CO_(2)`. Calculate the volume of `O_(2)` and `CO_(2)` at 1 atmosphere and body temperature of `37^(@)C` assuming 10 L of blood in human body.

The circulation of blood in human body supplies `O_(2)` and releases `CO_(2)`. The concentration of `O_(2)` and `CO_(2)` is variable but on the average, 100 mL blood contains 0.02 g of `O_(2)` and 0.08 g of `CO_(2)`. Calculate the volume of `O_(2)` and `CO_(2)` at 1 atmosphere and body temperature of `37^(@)C` assuming 10 L of blood in human body.
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  1. 100 mL blood contains `0.02 g O_(2)` and `0.08 g CO_(2)`
    `:.` 10 L=10,000 mL blood will contain `O_(2)=(0.02)/(100)xx10000=2g” and ” CO_(2)=(0.08)/(100)xx10000=8g`
    `PV=nRT=(w)/(M)RT`
    `:. ” For “O_(2),” “1xxV=(2)/(32)xx0.0821xx310 ” or “V_(o_(2))=1.59 L`
    For `CO_(2), ” ” 1xxV=(8)/(44)xx0.0821xx310 ” or “V_(CO_(2))=4.62 L`

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