The gas equation for one mole of oxygen relates its pressure, P (in atmospheres)

Question

The gas equation for one mole of oxygen relates its pressure, P (in atmospheres), its temperature, T (in K) and its volume, V (in cubic decimeters, dm^3) T = 16.574 middot 1/V - 0.52754 middot 1/V^2 - 0.3879 P + 12.187 VP. Find the temperature T and differential dT if the volume is 40 dm^3 and the pressure is 0.5 atmosphere. T = dT = Use your answer to part (a) to estimate how much the volume would have to change if the pressure increased by 0.2 atmosphere and the temperature remained constant. change in volume =

Explanation / Answer

(a) Given equation

T = 16.574 (1/V) - 0.52754 (1/V2) -0.3879 P + 12.187 VP

V= 40 dm3 , p= 0.5 atm , n=1

T = (16.574)( 1/40) - (0.52754) (1/402) - 0.3879 (0.5) + (40)(.5)

=-12.75

differentiating both sides

dT= 16.574 (-1/V2)dV - 0.52754 (-2/V3)dV -0.3879 dP + 12.187 (V dP + P dV)

(b) if dP =0.2, dT=0

0=(16.574)(-1/402)dV - 0.52754 (-2/403)dV - 0.3879 (.2)+ 12.187 ((40) (.2)+ (.5) dV)

dV = [(.3879 (.2) - (12.187)(40)(0.2)] / [(16.574) (-1/402)-(0.52754 )(-2/403)+ (12.187 )(40)(.5) ]

=-94.418 / 243.72

=-0.387

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