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Chapter 6: Problem 59

Which actions would you take to establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C} ?\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;\) (b) release \(0.80 \mathrm{g} \mathrm{O}_{2} ;\) (c) add \(2.00 \mathrm{g} \mathrm{He} ;\) (d) add \(0.60 \mathrm{g}\) He.

Short Answer

Expert verified
Based on this analysis, none of the proposed changes would establish exactly a pressure of 2.00 atm in the cylinder. However, option (a) would bring us closer to the target pressure of 2 atm, by increasing the current pressure of 0.5 atm to 1.0 atm.

Step by step solution

01

Identify Given Variables

This is what we know so far: we have a 2.24 L cylinder, initially filled with 1.60 g \(O_2\), and we want the pressure to become 2.00 atm. The temperature is \(0^{\circ}C\), but we will need to convert that into Kelvin before we use it in our Ideal Gas Law equations. So, the temperature in Kelvin is \(273 K\). Also, remember that the gas constant (R) for this scenario (where pressure is in atm and volume is in L) is \(0.0821 L•atm/mol•K\).
02

Calculating the Current Quantity of \(O_2\)

First, we need to calculate how many moles of \(O_2\) gas were initially in the cylinder. We can use the molar mass of \(O_2\) (which is 32.00 g/mol) to convert the given mass of 1.60 g into moles using the formula: \n \[n=\frac{mass}{molar~mass}\]\n Substituting the given values into the formula, \[n=\frac{1.60~g}{32.00~g/mol}\]\nwhich equals 0.05 mol. We'll use this value in our next calculation.
03

Initial Pressure Calculation

With all variables now at hand, the initial pressure in the cylinder can be calculated using the Ideal Gas Law equation. The rearranged formula for this calculation becomes: \n \[P=\frac{nRT}{V}\]\n Substituting the given values into the formula, \[P=\frac{0.05~mol~*~0.0821~L•atm/mol•K~*~273~K}{2.24~L}\] which yields a pressure of roughly 0.50 atm.
04

Analyze Each Scenario

Now that we have calculated the initial condition of the cylinder, we can move on to analyzing each case. It is important to remember that adding or removing gas from the cylinder will alter the number of moles (n) which makes pressure increase or decrease according to the Ideal Gas Law.
05

Analyzing Option (a)

In this case, another 1.60 g of \(O_2\) gas are added to the cylinder, which is equivalent to another 0.05 moles. So, the total moles of \(O_2\) gas in the cylinder is now 0.05mol + 0.05mol = 0.10 moles. Using the Ideal Gas Law, \[P=\frac{0.10~mol~*~0.0821~L•atm/mol•K~*~273~K}{2.24~L} =~ 1.00~ atm.\] This implies the pressure would become 1 atm, not the desired 2 atm.
06

Analyzing Option (b)

Here, 0.80 g, which is equivalent to 0.025 moles of \(O_2\) gas, is released from the cylinder. So, the total moles of \(O_2\) gas is now 0.05mol - 0.025mol = 0.025 moles. Using the Ideal Gas Law, \[P=\frac{0.025~mol~*~0.0821~L•atm/mol•K~*~273~K}{2.24~L} =~ 0.25~atm.\] This means the pressure will decrease to 0.25 atm, again not achieving 2 atm.
07

Analyzing Option (c)

Now 2.00 g of helium, equivalent to approximately 0.50 moles (using the molar mass of helium, which is 4.00 g/mol), is added to the cylinder. So, the total moles of gas in the cylinder is now 0.05mol (O2) 0.50mol (He) = 0.55 moles. Using the Ideal Gas Law, \[P=\frac{0.55~mol~*~0.0821~L•atm/mol•K~*~273~K}{2.24~L} =~ 6.73~atm.\] Under this scenario, the pressure in the cylinder would exceed the required 2 atm.
08

Analyzing Option (d)

Finally, if 0.60 g of helium (about 0.15 mol) is added to the cylinder, the total moles of gas in the cylinder is now 0.05mol (O2) 0.15mol (He) = 0.20 moles. Using the Ideal Gas Law, \[P=\frac{0.20~mol~*~0.0821~L•atm/mol•K~*~273~K}{2.24~L} =~ 2.45~atm.\] This situation will also result in a pressure above the desired 2 atm.

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Most popular questions from this chapter

Carbon monoxide, \(\mathrm{CO}\), and hydrogen react according to the equation below. $$3 \mathrm{CO}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of which reactant gas remains if \(12.0 \mathrm{LCO}(\mathrm{g})\) and \(25.0 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) are allowed to react? Assume that the volumes of both gases are measured at the same temperature and pressure.

A gaseous hydrocarbon weighing \(0.231 \mathrm{g}\) occupies a volume of \(102 \mathrm{mL}\) at \(23^{\circ} \mathrm{C}\) and \(749 \mathrm{mmHg} .\) What is the molar mass of this compound? What conclusion can you draw about its molecular formula?

We want to change the volume of a fixed amount of gas from \(725 \mathrm{mL}\) to 2.25 L while holding the temperature constant. To what value must we change the pressure if the initial pressure is \(105 \mathrm{kPa} ?\)

To establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C},\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;(\mathrm{b})\) add \(0.60 \mathrm{g} \mathrm{He}(\mathrm{g}) ;(\mathrm{c})\) add \(2.00 \mathrm{g} \mathrm{He}(\mathrm{g})\) (d) release \(0.80 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\)

A 0.168 L sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(26^{\circ} \mathrm{C}\) and a barometric pressure of \(737 \mathrm{mm} \mathrm{Hg}\). In the gas that is collected, what is the percent water vapor (a) by volume; (b) by number of molecules; (c) by mass? (Vapor pressure of water at \(26^{\circ} \mathrm{C}=25.2 \mathrm{mmHg}\).)
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