Question

The respiration of a suspension of yeast cells was measured by determining the decrease in pressure of the gas above the cell suspension. The apparatus was arranged so that the gas was confined to a constant volume, 16 $c{m}^3$ and the entire pressure change was caused by the uptake of oxygen by the cells. The pressure was measured in a monometer, the fluid of which had a density of 1.034 $g/c{m}^3$. The entire apparatus was immersed in a thermostat at $37$. In a $30$ minute observation period the fluid in the open side of the manometer dropped $37mm$. Neglecting the solubility of oxygen in the yeast suspension, compute the rate oxygen consumption by the cells in $m{m}^3$ of $O_2$ (STP) per hour.

Answer 1

As we know,
${\rho }_1{h}_1={\rho }_2{h}_2$
In one hour ($5\times 920h_1=37\times 2=74mm$)
$1.034\times 74=13.6\times h_2$
$h_2=mm$ of Hg
$P=\frac{5.626}{760}\times n\times 0.821\times 310$
$n=4.653\times {10}^{-3}mole/hr$
rate of $O_2$ consumption $=n\times 22400c{m}^3$
$=0.104c{m}^3/m$
$\left(O_{2}948399\right)=104m{m}^3/hr$