The greatest pressure of the following is that exerted by:
(a) a column of Hg 75.0 cm high (d=13.6 g/cm^3)
(b) 10.0 g H2 gas at STP
(c) a column of air 10 miles high; or
(d) a column of CCl4 60.0 cm high (d=1.59 g/cm^3)
The answer is b. I know that pressure=force/area and pressure can be atm or torr or bar. (c) is 1013 mb. Not sure the reasoning process here with the rest and how to compare with each other. thank you.
a) is 75/78 = <1 atm
b) If the H2 gas is at STP then P MUST be 1 atm.
c) An air column to infinity is 1 atm so 10 miles high will be <1atm. (1013 mb is 0.99 atm)
d)Here is a link to calculate P of CCl4 column. It is < 1 atm.
(Broken Link Removed) b must the right answer.
To compare the pressures exerted in each scenario, we need to calculate the pressure individually for each option.
Option (a) is the pressure exerted by a column of mercury. The density of mercury is given as 13.6 g/cm^3 and the height of the column is 75.0 cm. To calculate the pressure, we can use the formula P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height of the column. Plugging in the values, we have:
P(a) = (13.6 g/cm^3) * (9.8 m/s^2) * (0.75 m)
To convert grams to kilograms, we divide by 1000, and to convert centimeters to meters, we divide by 100:
P(a) = (13.6 kg/m^3) * (9.8 m/s^2) * (0.75 m)
Simplifying this expression further will give us the pressure exerted by the column of mercury in pascals (Pa).
Option (b) is the pressure exerted by 10.0 grams of hydrogen gas at STP (Standard Temperature and Pressure). To calculate the pressure, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
At STP, the temperature is 273.15 K, and the volume of 10.0 grams of hydrogen gas can be calculated using its molar mass. Then, we can rearrange the ideal gas law to solve for pressure P:
P(b) = (nRT) / V
Substituting the given values, we can calculate the pressure exerted by 10.0 grams of hydrogen gas.
Option (c) is the pressure exerted by a column of air 10 miles high. The pressure exerted by the atmosphere decreases as we move higher up. At sea level, the standard atmospheric pressure is 1 atmosphere (atm), which is approximately equal to 1013 millibars (mb) or 760 mmHg or torr.
Since the pressure decreases with height, we need to account for the change in pressure. One common approximation is that the pressure decreases by around 1 atmosphere for every 10 kilometers of altitude. By converting 10 miles to kilometers, we can calculate the decrease in pressure and subtract it from the standard atmospheric pressure to find the pressure at a height of 10 miles.
Option (d) is the pressure exerted by a column of carbon tetrachloride (CCl4). The density of CCl4 is given as 1.59 g/cm^3, and the height of the column is 60.0 cm. Similar to option (a), we can use the formula P = ρgh to calculate the pressure exerted by the column of CCl4.
Once you have calculated the pressure for each option, you can compare them to determine which one has the greatest pressure.