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## Q. 4.P.12

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.

## Verified Solution

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

$\int v d P+\int g d z=0$

$v=v_1\left(P / P_1\right)$

and:                     $P_1 v_1 \int \frac{ d P}{P}+\int g d z=0$

and:                     $P_1 v_1 \ln \left(P_2 / P_1\right)+g\left(z_2-z_1\right)=0$

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

∴                  $98,400 \times 0.841 \ln \left(P_2 / 98.4\right)+9.81(3000-0)=0$

and:                       $P_2=68.95 kN / m ^2$

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

$\log P=a+b T$

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

∴                         $\log P=-5.773+0.0209 T$

When $P_2=68.95, T=\underline{\underline{364 K }}$.