Question

The respiration of the suspension of yeast cells was measured by observing the decrease in pressure of gas above the cell suspension. The apparatus show that the gas was confined to a constant volume, 16 ml and the entire pressure change was caused by uptake of O2 by the cells. The pressure was measured in a manometer, the fluid of which had a density of I .034 gm/ml. The entire apparatus was immersed in a thermostat at 37°C. In a 30 min. observation period the fluid in open side of manometer dropped 37 mm. neglecting the solubility of O2 in yeast suspension, compute rate of O2 consumed by the cells in It of O2 (S.T.P.) per hour.

Answer 1

Ans > Density is 1.034 g/mL. The decrease in pressure is = density x g x height = 1034 x 9.8 x 0.037= 374.94 Pa Now PV= nRT. Let the initial pressure be P1 and final be P2. The initial moles be n1 and final be n2. Now P1-P2 = (n1-n2)RT/V = 374.94 => n1-n2 = {(374.94)(0.016)}/(8.314 x 310) = 0.00232 moles In one hour number of moles consumed would be 0.00464 moles. At STP the pressure and temperature would be different but the change in pressure is the same.The volume would remain same as the previously confined volume of 16mL. So at STP the number of moles of O₂ consumed would be = {374.94 x 0.016}/{8.314 x 273.15} = 0.00264 moles In one hour the amount of O₂ consumed is 0.00528 moles . Therefore the O₂ consumption rate is 0.00528 moles per hour.