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Chapter 11: Problem 24

The pressure in a tank of oxygen is \(2000.5 \mathrm{lb} / \mathrm{in} .{ }^{2}\) \(\left[760.00 \mathrm{~mm} \mathrm{Hg}=14.696 \mathrm{lb} / \mathrm{in}^{2}\right]\) (a) How many millimeters of mercury is this? (b) How many atmospheres is this?

Short Answer

Expert verified
(a) The pressure in the tank of oxygen is approximately \(103398.8 \, \text{mm Hg}\). (b) The pressure in the tank of oxygen is approximately \(136.15 \, \text{atm}\).

Step by step solution

01

Convert lb/in² to mm Hg

To convert lb/in² to mm Hg, we will use the given conversion factor 760 mm Hg = 14.696 lb/in². We set up the conversion as follows: \( \frac{2000.5 \, \text{lb/in}^{2}}{1} \cdot \frac{760 \, \text{mm Hg}}{14.696 \, \text{lb/in}^{2}} \)
02

Calculate the pressure in mm Hg

Next, we multiply and divide the numbers: \( \frac{2000.5 \cdot 760}{14.696} \) which equals approximately 103398.8 mm Hg. So the pressure in the tank is roughly 103398.8 mm Hg.
03

Convert lb/in² to atm

Now we will convert the pressure from lb/in² to atm, using the given conversion factor 1 atm = 14.696 lb/in². We set up the conversion as follows: \( \frac{2000.5 \, \text{lb/in}^{2}}{1} \cdot \frac{1 \, \text{atm}}{14.696 \, \text{lb/in}^{2}} \)
04

Calculate the pressure in atm

Multiply and divide the numbers: \( \frac{2000.5}{14.696} \) which equals approximately 136.15 atm. So the pressure in the tank is approximately 136.15 atm. To sum up, the pressure in the tank of oxygen is: (a) about 103398.8 mm Hg, and (b) about 136.15 atm.

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Most popular questions from this chapter

A gas tank contains \(\mathrm{CO}_{2}\) at a pressure of \(6.80 \mathrm{~atm}\). What would the \(\mathrm{CO}_{2}\) pressure be if the container were (a) twice as large and (b) one-fourth as large?

A weather forecaster reports the barometric pressure as \(29.7\) inches of mercury. \([1 \mathrm{in.}=2.54 \mathrm{~cm}]\) (a) How many millimeters of mercury is this? (b) How many atmospheres is this?

A tank initially at a pressure of \(6.70\) atm contains \(20.0\) moles of oxygen gas at \(25.0^{\circ} \mathrm{C}\). Its temperature is then increased to \(45.0{ }^{\circ} \mathrm{C}\), and an additional \(10.0\) moles of gas are added. What is the final pressure in the tank?

A 5.00-L container is filled with \(0.8004\) g of helium gas at \(107.0^{\circ} \mathrm{C}\). What is the pressure of the gas in atmospheres and in millimeters of mercury?

Consider two identical \(1-\mathrm{L}\) containers, both at room temperature \((300 \mathrm{~K})\). One of them contains 1 mole of helium gas, and the other contains 1 mole of hydrogen gas. Is the pressure higher in the helium container, higher in the hydrogen container, or the same in the two containers?
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