Question

A patient Vasu had to survive for a month on artificial respiration before he passed away. A nurse was instructed to give him 7.5g of glucose per hour for consumption to meet the energy requirement. What volume of carbogen ( 95% O₂ + 5% CO₂) had to be supplied for Vasu assuming that he was kept under the conditions of 27⁰C and normal atmospheric pressure?

Answer 1

Amount of glucose instructed to give him per hour = 7.5g

Amount of glucose instructed to give him per day = 7.5g ×24

= 180g

Amount of glucose instructed to give him per month = 180×30

= 5400g

Molar mass of glucose (C₆H₁₂O₆) = 6 ×12 + 12 ×1 + 6×16

= 180g

Number of moles of glucose =$\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}$

= $\frac{5400}{180}$

= 30

C₆H₁₂O₆ (s)+6O₂ (g)→6CO₂ (g) +6H₂O(l)

Volume of 1 mole of any substance at STP = 22.4 L

1 molecule of glucose requires 6 molecules of O₂ for combustion.

Volume of O₂ required for the combustion of 1 mole of glucose at STP = 6 ×22.4L

= 134.4 L

Total Volume of O₂ required for the combustion at STP at 30 days = 134.4 ×30

= 4032 L

Given that

Initial pressure P₁= 1 atm

Final Pressure P₂ = 1 atm

Initial temperature T₁ = 0^oC = 273 K

Final temperature T₂= 27^oC

= 27 + 273 K

= 300 K

Initial volume V_{1 =} 4032 L

Final volume V_{2 =} ?

According to gas law,

$\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

${\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{T}}_2}{{\mathrm{T}}_1{\mathrm{P}}_2}$

${\mathrm{V}}_2=\frac{1\times 4032\times 300}{273\times 1}$

V_{2 =} 4430.76 L

Oxygen required is 4430.76 L.

95% of oxygen is present in 100% of carbogen.

That is 95L of oxygen is present in 100L of carbogen.

1L of oxygen is present in 100L of carbogen = $\frac{100}{95}$

4430.76 L of oxygen present in =$\frac{100}{95}\times 4430.76$

= 4663.95 L of carbogen.