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(a)Pressure decreases as oneascends the atmosphere . If the density of air is rho ,what is the change in pressure dp over differential height dh ? (b)Considering the pressure P to be proportional to the density find the pressure P at a height h if the pressure on the surface of the earth is P_(o).(c ) If P_(o)=1.03xx10^(5)N//m^(-2),rho_(o)=1.29kg//m^(3)andg=9.8m//s^(2) what height will the pressure frop to (1)/(10) the value at the surface of earth ? (d) This model of the atmosphere works for relatively small distance .Identify the underlying assumption that limits the model. |
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Answer» SOLUTION :Since the air is less dense at higher up and hence pressure is also less. (a) Consider a HORIZONTAL portion of air with cross section A and height dh.  Let the pressure on the top surface and bottom surface be P and `P_(1)+dP`. If the portion is in equilibrium then the net upward force must be balanced by the weight. `(P+dP)A-PA=-mg"" (because"mass"="volume"xx"density")` `therefore(dP)A=-rho(Adh)g` `(rho=` density of air) `thereforedp=-pgdh`...(1) Negative sign indicates that pressure decreases at height increases . (b)Let the density of air on the EARTH.s surface be `rho_(o)`, then Pressure `prop` density `thereforePproprhoandP_(o)proprho_(o)` `therefore(P)/(P_(o))=(rho)/(rho_(o))` `thereforerho=((P)/(P_(o)))rho_(o)`....(2) PUTTING value of equation (2) in (1), `thereforedp=-((P)/(10))rho_(o)dgh` `therefore(dP)/(P)=-(rho_(o)g)/(P_(o))dh` `thereforeh=(P_(o))/(rho_(o)g)ln(10)` `thereforeh=(P_(o))/(rho_(o)g)[2.363log_(10)^(10)]` `=((1.093xx10^(5)))/(1.22xx9.8)xx2.363(1)=0.16xx10^(5)m` `thereforeh=16xx10^(3)m` (d)Pressure `Pproprho` (For isothermal prosess)temperature remains constant only NEAR the surface of earth at greater height relation `Pproprho` does not obay.Hence ,for small distances near earth `Pproprho`is obay, this is not for greater distances . This is the limitation of this model . |
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