Oxygen for patients is kept in tanks at a pressure of 6.6 * 10^6 Pa and a tempar

Question

Oxygen for patients is kept in tanks at a pressure of 6.6 * 10^6 Pa and a temparature of 300 k What volume does 4.0 m ^2 of gas in the tank occupy in the room What is the average kinetic energy per oxygen molecule in the potients room if the patient takes a full breath and holds the oxygen in his lungs. it increases in temparature by 6 % How does this change the pressure in the lungs Shows a p^V diagram of the air in the human lung when a patient inhales and exhales the oxygen. Approximately how much work is done during this process these are 7.5 * 10 ^-3 mm Hg in 1 Pa

Explanation / Answer

(A) The process is isothermal Hence using Boyle's Law P1*V1  = P2*V2

so Volume of gas in room (V2) = 6.6*10^6*4/10^5 = 264 m^3

(B) Average kinetic energy is given by (K.Eav) = 3/2**T where is the Boltzman constant

Hence   K.Eav = 5/2*1.38*10^(-23)*300 = 6.2*10^(-21) J

(C) Here the process is isochoric hence pressure is directly proportional to temperature. So pressure will also increase by 6% (3rd option)

(C) Work done in a P-V diagram is just the area of the closed loop. After calculating the area by breaking the quadrilateral roughly into two triangles and on parallelogram we get the area 7.6 mm Hg -L

Which is cloose to the work done value of 1 J

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