A 1.20 kg ball is at rest on the floor in a 2.10 m times 2.10 m times 2.10 m roo

Question

A 1.20 kg ball is at rest on the floor in a 2.10 m times 2.10 m times 2.10 m room of air at STP. Air is 80% nitrogen (N_2) and 20% oxygen (O_2) by volume. What is the thermal energy of the air in the room? Express your answer with the appropriate units. What fraction of the thermal energy would have to be conveyed to the ball for it to be spontaneously launched to a height of 1.50 m? By how much would the air temperature have to decrease to launch the ball? Express your answer with the appropriate units.

Explanation / Answer

Given dimensions of room = 2.1*2.1*2.1m^3 = 9.261 m^3

A. For air at STP
   1 mole air occupies 22.4 lit = 22.4*10^-3 m^3
   so number of moles of air in the room , n = 9.261/22.4*10^-3 = 413.4375
   out of this 80 p.c. is N2 and 20 P.c. is O2
   so number of moles of N2, n1 = 0.8*n = 330.75
   so number of moles of O2, n2 = 0.2*n = 82.6875
   Also, heat capacity at constant volume is Cv
   Cv for N2 = 0.743
   Cv for O2 = 0.658

   so thermal energy in air = E = [n1Cv1 + n2*Cv2]*T
   T = 273.15 K
   hence E = [330.75*0.743 + 82.68*0.658]*273.15 = 82165.89 J
B. mass of ball, m = 1.2 kg
   energy required to reach h = 1.5 m = mgh
   PE = 1.2*9.81*1.5 = 17.658 J
   Fraction in terms of E = PE/E = 2.149*10^-4

C. let temperature change in air be dT for this process then
   [330.75*0.743 + 82.68*0.658]*dT = 17.658
   or dT = 0.0588 K

We have assumed Cv stays constant with fall in temperature

   

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