The gauge pressure inside a vessel is - 40kPa, at an elevation of 5000m. a) What is the absolute pressure? b) At this elevation, what temperature does water boil?

a) Pabs = 48960 KPa

b) T = 433.332 °C

Explanation:

Pabs = Pgauge + d*g * h

∴ d = 1000 Kg/m³

∴ g = 9.8 m/s²

∴ h = 5000 m

∴ P gauge = - 40 KPa * (1000 Pa / KPa) = - 40000 Pa; Pa≡Kg/m*s²

⇒ Pabs = - 40000 Kg/ms² + (1000 Kg/m³ * 9.8 m/s² * 5000 m)

⇒ Pabs = 48960000 Pa = 48960 KPa

a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:

P (T) = 0.61094 exp (17.625*T / (T + 243.04)) ... P (KPa) ∧ T (°C) ... from literature

∴ P = 48960 KPa

⇒ (48960 KPa / 0.61094) = exp (17.625T / (T + 243.04))

⇒ 80138.803 = exp (17.625T / (T + 243.04))

⇒ Ln (80138.803) = 17.625T / (T + 243.04))

⇒ 11.292 * (T + 243.04) = 17.625T

⇒ 11.292T + 2744.289 = 17.625T

⇒ 2744.289 = 17.625T - 11.292T

⇒ 2744.289 = 6.333T

⇒ T = 433.332 °C