Question

18. Consider a sealed piston containing a mixture of 22.0% O2 and 78.0% N2. The piston is pushed down and the volume of the gas decreases from 1.42 L to 1.02 L. Assuming that temperature is constant during this process, how does the change in volume affect the partial pressure of O2 in the container? A. B. C. The partial pressure of O2 increases by 6.20%. The partial pressure of O2 increases by 8.62%. The partial pressure of O2 increases by 28.2%. The partial pressure of O2 increases by 39.2%. The partial pressure of O2 increases by 71.8%. E.

Answer 1

Answer:-

Firstly using the ideal gas law at constant temperature, partial pressure of oxygen is calculated then using the given data, partial pressure of oxygen is calculated.

The answer is given in the image,

Answers Vi=1.421 P, EP V2 = 1.02 P2=? Pozcu) = Px0.22=0.222 At constant, P, V, = P₂ V2 P2== PX1.42 7.02 P2=1-392P Parcg) = 1.39f0-22=0.3063P Increase in poz=0.30638- 0.22f83) 50.0863P Yo increase = 0.0863 x100 = 39.2°/- 0-22 50, D. The partial pressure of Oz is increased by 39-2%.