14.4: Problem 13 Previous Problem Problem List Next Problem (1 point) The gas eq

Question

14.4: Problem 13 Previous Problem Problem List Next Problem (1 point) The gas equation for one mole of oxygen relates its pressure, P (in atmospheres), its temperature, T (n K) , and its volume, V (in cubic decimeters, dm3): T 16.574 0.52754 -0.3879 P + 12.187 VP (a) Find the temperature T and differential dT if the volume is 20 dm1 and the pressure is 1 atmosphere. T= dT= (b) Use your answer to part (a) to estimate how much the volume would have to change if the pressure increased by 0.1 atmosphere and the temperature rema change in volume = | ned constant.

Explanation / Answer

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

V= 20 dm3

P= 1 atm

T= 16.574/20 - 0.52754/202-0.3879 x 1+12.187x 20 x1

T= 0.8287-0.00131885-0.3879+243.74

T= 244.179 K

dT= (dT/dP)dP + (dT/dV)dV

dT= (-0.3879+12.187V)dP+(-16.574/V2+1.05508/V3+12.187P)dV

dT= (-0.3879+12.187 X 20)dP+(-16.574/202+1.05508/203+12.187X 1)dV

dT= (243.35)dP+(-0.041435+0.000131885+12.187)dV

dT= 243.35dP+12.145dV

b) Pressure increases by 0.1 atm

Constant temperature , dT=0

dT= 243.35dP+12.145dV

0=243.35 x 0.1+12.145dV

-24.335= 12.145dV

dV= -24.335/12.145= -2.0037

Volume decreases by 2.0037 dm3

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