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gases are different from solids and liquids. in a sample of gas, the mo…

Question

gases are different from solids and liquids. in a sample of gas, the molecules are far apart. the gas molecules also move around and collide with each other as well as with the walls of the container. these collisions generate pressure. the pressure of a gas can be measured in different units. one convenient unit of measure is called the atmosphere (atm) because it is based on atmospheric pressure. at sea level, the average pressure is 1 atm. as you get higher in altitude, the pressure steadily drops until you leave the atmosphere, where the pressure is very close to 0 atm.
the table below shows the different commonly used units of measuring gas pressure. use this table in the pressure unit conversions.

unitabbreviation1 atm equivalent
millimeters of mercurymmhg760 mmhg
torrtorr760 torr
inches of mercuryin. hg29.9 in. hg
pounds per square inch (psi)lb/in.²14.7 lb/in.²
pascalpa101,325 pa

part b
the pressure in car tires is often measured in pounds per square inch (lb/in.²), with the recommended pressure being in the range of 25 to 45 lb/in.². suppose a tire has a pressure of 28.5 lb/in.². convert 28.5 lb/in.² to its equivalent in atmospheres.
express the pressure numerically in atmospheres.
view available hints)
28.5 lb/in.² = \boxed{?} atm

part c
high - pressure liquid chromatography (hplc) is a method used in chemistry and biochemistry to purify chemical substances. the pressures used in this procedure range from around 500 kilopascals (500,000 pa) to about 60,000 kpa (60,000,000 pa). it is often convenient to know the pressure in torr. if an hplc procedure is running at a pressure of 3.64×10⁸ pa, what is its running pressure in torr?

Explanation:

Step1: Identify conversion factor

From the table, \(1\ \text{atm} = 14.7\ \text{lb/in}^2\). So the conversion factor is \(\frac{1\ \text{atm}}{14.7\ \text{lb/in}^2}\).

Step2: Apply conversion

Multiply the given pressure (\(28.5\ \text{lb/in}^2\)) by the conversion factor:
\(28.5\ \text{lb/in}^2 \times \frac{1\ \text{atm}}{14.7\ \text{lb/in}^2}\)
Simplify: \(\frac{28.5}{14.7} \approx 1.9388\)

Answer:

\(1.94\) (rounded to two decimal places, or \(1.939\) for more precision)