Question

Pressure Drop with Altitude Pressure (atm) 0 2000 4000 10000 12000 14000 6000 8000 Height (meters)

Answer 1

Using the known data from the plot and with the help of spreadsheet I have determined the pressure at Lukla, namche bazaar and Everest base camp.

i am sharing the picture of spreadsheet.

further calculations

boiling of water refers to following heterogeneous equilibrium

H2O(1) = H2O(0)


Here, k, = PHO
(partial pressure of water vapour) and we know at boiling point partial pressure of water vapour = atmospheric pressure. So, at lukla $P_{H_2O}$ = 0.745 atm, at namche it is 0.7028 atm and at Everest base camp it is 0.529 atm.

Applying thermodynamics concept,

$\Delta G^0 = \Delta H^0 - T \Delta S^0$and $\Delta G^0 = - 2.303 \times RT \times log K_p$

thus, $- 2.303 \times RT \times log K_p = \Delta H^0 - T \Delta S^0$ here T refers to boiling point . We know the value of $K_p , \Delta H^0 and \Delta S^0$ so T can be calculated in each case.

first , sea level kp = 1 atm (given), so applying above formula

$- 2.303 \times RT \times log (1) = \Delta H^0 - T \Delta S^0$

$0 = \Delta H^0 - T \Delta S^0 thus T = \frac{\Delta H^0}{\Delta S^0}$

$T = \frac{44100}{118.20}$ = 373.09 K

similarly for lukla

kp = 0.745 (determined), $- 2.303 \times RT\times log (0.745) = \Delta H^0 - T \Delta S^0$

= $- 2.303 \times 8.314 \times T \times (-0.12784) = 44100- T \times 118.2$ (R = 8.314 J/ mole/ K)

on solving above equation, T = 365.52 K

for namche

kp = 0.7028 (determined), $- 2.303 \times RT\times log (0.7028)= \Delta H^0 - T \Delta S^0$

= $- 2.303 \times 8.314 \times T \times (-0.153) = 44100- T \times 118.2$ (R = 8.314 J/ mole/ K)

on solving above equation, T = 364 K

for Everest base camp

kp = 0.5298 (determined), $- 2.303 \times RT\times log (0.5298) = \Delta H^0 - T \Delta S^0$

= $- 2.303 \times 8.314 \times T \times (-0.2758) = 44100- T \times 118.2$ (R = 8.314 J/ mole/ K)

on solving above equation, T = 357.14 K.

Thus boiling point at various places

sea level = 373.09 K

lukla = 365.52 K

namche = 364 K

base camp = 357.14 k

conversion factor for Kelvin(K) and degree Celsius is