Show that the atm.pressure of 1.013x10^5Pa will support (a)10.3m of water (b)760mmHg of Hg.If the density of water is 10^3kg/m3 and density of mercury is 13.6x10^3kg/m3.
water mass = 1000 kg/m^3 * volume
volume of 10.3 high by 1 meter square base = 10.3 m^3
so mass of water sitting on that square =
10.3*10^3 kg
weight = m g = 9.81*10.3*10^3
= 101 *10^3 = 1.01*10^5 N/m^2 or Pascals
do the HG the same way.
To determine whether the atmospheric pressure of 1.013x10^5 Pa can support a certain height of water or mercury, we need to compare the pressure exerted by the column of liquid to the atmospheric pressure.
(a) To find if the atmospheric pressure can support a height of 10.3 m of water, we can use the hydrostatic pressure formula:
Pressure = density * gravity * height
Given:
Density of water (ρ) = 10^3 kg/m^3
Acceleration due to gravity (g) = 9.8 m/s^2
Height of water column (h) = 10.3 m
Using the formula:
Pressure = ρ * g * h
Substituting the values:
Pressure = (10^3 kg/m^3) * (9.8 m/s^2) * (10.3 m)
Calculating:
Pressure = 10^3 * 9.8 * 10.3 Pa
Comparing the pressure to atmospheric pressure:
Since the atmospheric pressure is 1.013x10^5 Pa, and the pressure exerted by the height of water is considerably smaller (around 1.003x10^5 Pa), we can conclude that the atmospheric pressure can support a height of 10.3 m of water.
(b) Similarly, to find if the atmospheric pressure can support 760 mmHg of Hg, we need to convert the given pressure to Pascal (Pa).
Given:
Density of mercury (ρ) = 13.6x10^3 kg/m^3
Height of mercury column (h) = 760 mmHg
To convert mmHg to Pa:
1 mmHg = 133.322 Pa
Pressure = (760 mmHg) * (133.322 Pa/mmHg)
Calculating:
Pressure = 101325.12 Pa
Comparing the pressure to atmospheric pressure:
Since the atmospheric pressure is 1.013x10^5 Pa, and the pressure exerted by the height of mercury is considerably smaller (around 1.01325x10^5 Pa), we can conclude that the atmospheric pressure can support 760 mmHg of mercury.