If temperature does not change with height, estimate the boiling point of water at a height of 3000 m above sea-level. The barometer reading at sea-level is 98.4 kN/m² and the temperature is 288.7 K. The vapour pressure of water at 288.7 K is 1.77 kN/m². The effective molecular weight of air is

Question

If temperature does not change with height, estimate the boiling point of water at a height of 3000 m above sea-level. The barometer reading at sea-level is 98.4 kN/m² and the temperature is 288.7 K. The vapour pressure of water at 288.7 K is 1.77 kN/m². The effective molecular weight of air is 29 kg/kmol.

Accepted Answer

The air pressure at 3000 m is [latex]P_2[/latex] and the pressure at sea level, [latex]P_1=98.4 kN / m ^2[/latex].

[latex]\int v d P+\int g d z=0[/latex]

[latex]v=v_1\left(P / P_1 ight)[/latex]

and: [latex]P_1 v_1 \int \frac{ d P}{P}+\int g d z=0[/latex]

and: [latex]P_1 v_1 \ln \left(P_2 / P_1 ight)+g\left(z_2-z_1 ight)=0[/latex]

[latex]v_1=(22.4 / 29)(288.7 / 273)(101.3 / 98.4)=0.841 m ^3 / kg[/latex].

∴ [latex]98,400 imes 0.841 \ln \left(P_2 / 98.4 ight)+9.81(3000-0)=0[/latex]

and: [latex]P_2=68.95 kN / m ^2[/latex]

The relationship between vapour pressure and temperature may be expressed as:

[latex]\log P=a+b T[/latex]

When, [latex]T=288.7, P=1.77 kN / m ^2[/latex]
and when, [latex]T=373, P=101.3 kN / m ^2[/latex]

∴ [latex]\log P=-5.773+0.0209 T[/latex]

When [latex]P_2=68.95, T=\underline{\underline{364 K }}[/latex].

Q. 4.P.12

Chapter 4

Q. 4.P.12

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