question_answer 52)

                    (a) Pressure decreases as one ascends the atmosphere. If the density of air is \[\rho \], what is the change in pressure \[dp\] over a 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 \[{{p}_{0}}\].                 (c) If \[{{\rho }_{0}}=1.03\times {{10}^{3}}\,N\,{{m}^{s}},\,\,{{\rho }_{0}}=1.2\,kg\,{{m}^{-3}}\] and \[g=9.8\,m\,{{s}^{-2}},\] at what height will the pressure drop to (1/10) the value at the surface of the earth?                 (d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.                

Answer:

                  (a) Refer Law of atmosphere (Additional Topic) in chapter 10(A) of New Millennium Physics, Class XI.                 From eqn. (i),                 \[\frac{dp}{dy}=-\rho g,\,\,dp=-\rho gdh\]                    (replacing \[dy\] by \[dh\])                 (b) Since \[p\propto \rho ,\,\frac{p}{{{p}_{0}}}=\frac{\rho }{{{\rho }_{0}}}\] or \[\rho =\frac{{{\rho }_{0}}}{{{p}_{0}}}p\]                 Clearly, \[dp=-\frac{{{\rho }_{0}}g}{{{p}_{0}}}pdh\] or \[\frac{dp}{p}=-\frac{{{\rho }_{0}}g}{{{p}_{0}}}dh\]                 On integration, \[\int\limits_{{{p}_{0}}}^{p}{\frac{dp}{p}}=-\frac{{{\rho }_{0}}g}{{{p}_{0}}}\int\limits_{0}^{h}{dh}\]                 or \[\ln \frac{p}{{{p}_{0}}}=-\frac{{{\rho }_{0}}g}{{{p}_{0}}}h\]   ? (i)                 whence, \[p={{p}_{0}}{{e}^{-({{\rho }_{0}}g/{{p}_{0}})h}}={{p}_{0}}\exp \left( -\frac{{{\rho }_{0}}g}{{{p}_{0}}}h \right)\]                 (c) When \[p=\frac{{{p}_{0}}}{10},\,\frac{p}{{{p}_{0}}}=\frac{1}{10}\] and \[h={{h}_{0}}\]             From eqn. (i), \[\ln \frac{1}{10}=-\frac{{{\rho }_{0}}g}{{{p}_{0}}}\]             or \[{{h}_{0}}=-\frac{{{p}_{0}}}{{{\rho }_{0}}g}\ln \frac{1}{10}=-\frac{{{p}_{0}}}{{{\rho }_{0}}g}(-2.3026)\]                 [as \[\ln \frac{1}{10}=\ln \,1-\ln \,10=0-2.3026\,\log \,10\]\[=-2.3026\]]                 or \[{{h}_{0}}=\frac{(1.013\times {{10}^{5}}\,N/{{m}^{2}})}{(1.29\,kg/{{m}^{3}})(9.8\,m/{{s}^{2}})}(2.3026)\]             \[=0.1845\times {{10}^{5}}\,m\]             \[=18.45\,km\]                 (d) We have assumed that \[p\propto \rho \] (true only for the isothermal case) which is valid only for relatively small distances.