Explanation The statement of Barometric distribution law is: The pressure of a gas, which behave like ideal, decreases with the increment of height under gravitational field. We assume a long column having height, h of a gas under the gravitational field of strength, g. The pressure of the gas at the bottom of the column i.e. h=0, the pressure is P 0 and at height, h the pressure is P. If we consider a small slice, of the column at height, h, where the thickness of the height is and as per the law the decrease pressure is -dp. The gas has density at height, h and gravitational field, g. Pressure(p)=Density(ρ) × Gravitational field(g) × Height(h), so, -dp = ρ ×g × dh...................Equation 1 (Where is density of the gas). As per ideal gas equation, PV = nRT (Where P, V, n, R and T is the pressure, volume, number of moles of gas, universal gas constant and temperature respectively. Number of moles of gas ( n ) = Weight of the gas ( W ) Molecular mass of the gas ( M ) PV = W M RT P × M = W V RT So, P × M = ρRT ( As , W M = ρ ) ρ = P × M R × T Now putting the expression of in equation 1, we get - dp = P × M R × T × g × dh - dP P = M × g R × T dh Now integrating both side - ∫ dP P = M × g R × T ∫ dh - 1 nP = M × g R × T × h + Z (where,Z is the integrating constant) When h=0, then P=P 0 , thus -1n P 0 = Z - 1 nP = M × g R × T × h - 1 nP 0

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Chapter 10, Problem 30E

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To determine: Pressure of the gas sample in each case

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Pressure of the gas sample in each case

Solution Summary: The author explains the Barometric distribution law, which states that the pressure of a gas, behave like ideal, decreases with the increment of height under gravitational field.

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To determine: Pressure of the gas sample in each case

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Chapter 10 Solutions