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 = 16.574.1/v - 0.52754 . 0.3879 P + 12.187 V P. Find the temperature T| and differential dt| if the volume is 28 dm^3| and the pressure is 0.5 atmosphere. T =| dT =| Use your answer to part (a) to estimate how much the pressure would have to change if the volume increased by 1 dm^3| and the temperature remained constant.change in pressure =

Explanation / Answer

T = 16.574 * (1/v) - 0.52754 * (1/v2) - 0.3879 P + 12.187 V P

when V = 1 dm3 = 0.001 m3and P = 0.5 atm

T = (16.574 * (1/0.001)) - ( 0.52754 * (1/0.000001) ) - (0.3879 * 0.5 ) + ( 12.187 * 0.001 * 0.5 )

= (16.574 * 1000) - ( 0.52754 * 1000000) - (0.3879 * 0.5 ) + ( 12.187 * 0.001 * 0.5 )

=(16574 - 527540- 0.7758 +0.0060935 )

= -510966.7697065 degree cesius

=

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