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Chapter 11: Problem 80

A sample of argon at \(0.0^{\circ} \mathrm{C}\) and \(700.0 \mathrm{~mm} \mathrm{Hg}\) occupies \(15.0 \mathrm{~mL}\). What is its volume at STP?

Short Answer

Expert verified
The volume of the argon sample at STP conditions is \(13.9 \ \mathrm{mL}\).

Step by step solution

01

Convert the given information to appropriate units.

The initial temperature, \(T_1\), = 0.0°C. We need to convert this to Kelvin by adding 273.15: \(T_1\) = 0.0 + 273.15 = 273.15 K The initial pressure, \(P_1\), = 700.0 mm Hg. We need to convert this to atm by using the relationship 1 atm = 760 mm Hg: \(P_1\) = 700.0 mm Hg × \(\frac{1 \ \mathrm{atm}}{760 \ \mathrm{mm \ Hg}}\) = 0.9211 atm The initial volume, \(V_1\), is given as 15.0 mL. We will use this as-is.
02

Identify final temperature and pressure values.

We need to find the final volume of the argon sample at STP. The final temperature, \(T_2\), and final pressure, \(P_2\), at STP should be 273.15 K and 1 atm, respectively.
03

Substitute the known values into the Combined Gas Law formula and solve for the final volume, \(V_2\).

We will now substitute the known values into the Combined Gas Law formula and solve for the final volume, \(V_2\). \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) \(\frac{0.9211 \ \mathrm{atm} \times 15.0\ \mathrm{mL}}{273.15 \ \mathrm{K}} = \frac{1 \ \mathrm{atm} \times V_2}{273.15 \ \mathrm{K}}\) Next, we will solve for \(V_2\): \(V_2 = \frac{0.9211 \ \mathrm{atm} \times 15.0 \ \mathrm{mL} \times 273.15 \ \mathrm{K}}{273.15 \ \mathrm{K} \times 1 \ \mathrm{atm}} = 13.855 \ \mathrm{mL}\)
04

Round the final answer to a reasonable number of significant figures.

We will round the final volume to three significant figures, like the given pressure and volume values, giving us a final volume of: \(V_2 = 13.9 \ \mathrm{mL}\) So, the volume of the argon sample at STP conditions is 13.9 mL.

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Most popular questions from this chapter

When it is heated, potassium chlorate decomposes to potassium chloride and oxygen: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) A sample of solid \(\mathrm{KClO}_{3}\) decomposes to produce \(150.0 \mathrm{~mL}\) of oxygen at \(22.0^{\circ} \mathrm{C}\) and \(780.5 \mathrm{~mm} \mathrm{Hg}\). How many grams of \(\mathrm{KClO}_{3}\) decomposed?

What is the molar mass of a gas whose density is \(1.52 \mathrm{~g} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) and 1 atm pressure?

An automobile tire at \(22^{\circ} \mathrm{C}\) with an internal volume of \(20.0 \mathrm{~L}\) is filled with air to a total pressure of 30 psi (pounds per square inch). \(\left[1 \mathrm{~atm}=14.696 \mathrm{lb} / \mathrm{in} .^{2}\right]\) (a) What is the amount in moles of air in the tire? (b) If the air were entirely nitrogen \(\left(\mathrm{N}_{2}\right)\), how many grams of it would be in the tire? How many pounds of it would be in the tire? \([453.6 \mathrm{~g}=1 \mathrm{lb}]\)

A gas at \(25.0{ }^{\circ} \mathrm{C}\) occupies a volume of \(5.00\) gallons and exerts a pressure of \(755 \mathrm{~mm} \mathrm{Hg}\). Express (a) the temperature of the gas in Kelvin, (b) the volume of the gas in liters, and (c) the pressure of the gas in atmospheres.

A steel cylinder is filled with a gas. The initial pressure of this gas is \(5.00 \mathrm{~atm}\), and the initial temperature is \(200.00^{\circ} \mathrm{C}\). The cylinder is heated to a final temperature of \(400.00{ }^{\circ} \mathrm{C}\). It appears that the temperature has doubled, but the final pressure is not \(10.0 \mathrm{~atm}\). Why doesn't the pressure double as it did in the previous example? What is the final pressure? What valuable lesson does this question teach you?
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