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Chapter 6: Problem 10

An 886 mL sample of \(\mathrm{Ne}(\mathrm{g})\) is at \(752 \mathrm{mmHg}\) and \(26^{\circ} \mathrm{C}\). What will be the new volume if, with the pressure and amount of gas held constant, the temperature is (a) increased to \(98^{\circ} \mathrm{C} ;\) (b) lowered to \(-20^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The new volume when the temperature is increased to \(98^{\circ} \mathrm{C}\) is about 1094.24 mL, and the new volume when the temperature is lowered to -20^{\circ} \mathrm{C} is about 751.54 mL.

Step by step solution

01

Convert temperatures to Kelvin

Convert all Celsius temperatures to Kelvin since the gas laws are based on the absolute (Kelvin) temperature scale. You can do this by adding 273.15 to the Celsius temperature.\nSo, for the initial temperature: \(26^{\circ}\mathrm{C} + 273.15 = 299.15 \mathrm{K}\) \n(a) For the final temperature when increased to \(98^{\circ}\mathrm{C}\): \(98^{\circ}\mathrm{C} + 273.15 = 371.15 \mathrm{K}\) (b) For the final temperature when lowered to \(-20^{\circ}\mathrm{C}\): \(-20^{\circ}\mathrm{C} + 273.15 = 253.15 \mathrm{K}\)
02

Use Charles's Law to find the new volume

Use Charles's Law: \(V_1/T_1 = V_2/T_2\), where \(V_1\) and \(T_1\) denote the initial volume and temperature, and \(V_2\) and \(T_2\) denote the final volume and temperature. (a) Solve for the new volume \(V_2\) when the temperature is increased to \(98^{\circ}\mathrm{C} (or 371.15\mathrm{K}\): \((886\mathrm{mL}/299.15\mathrm{K}) = V_2/371.15\mathrm{K}\) \n\nThis yields \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 371.15\mathrm{K} \)\n (b) Solve for the new volume \(V_2\) when the temperature is lowered to \(-20^{\circ}\mathrm{C} (or 253.15\mathrm{K}\): \((886mL/299.15\mathrm{K}) = V_2/253.15\mathrm{K}\)\n\nThis yields \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 253.15\mathrm{K}\)
03

Calculate the new volumes

Perform the calculations in Step 2 to get the new volumes. (a) For the increased temperature: \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 371.15\mathrm{K} = 1094.24 \mathrm{mL}\) (rounded to two decimal places) (b) For the decreased temperature: \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 253.15\mathrm{K} = 751.54 \mathrm{mL}\) (rounded to two decimal places)

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Most popular questions from this chapter

Monochloroethylene is used to make polyvinylchloride (PVC). It has a density of \(2.56 \mathrm{g} / \mathrm{L}\) at \(22.8^{\circ} \mathrm{C}\) and 756 mmHg. What is the molar mass of monochloroethylene? What is the molar volume under these conditions?

A 1.65 g sample of \(\mathrm{Al}\) reacts with excess \(\mathrm{HCl}\), and the liberated \(\mathrm{H}_{2}\) is collected over water at \(25^{\circ} \mathrm{C}\) at a barometric pressure of \(744 \mathrm{mmHg} .\) What volume of gaseous mixture, in liters, is collected? $$2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})$$

A nitrogen molecule ( \(\mathrm{N}_{2}\) ) having the average kinetic energy at \(300 \mathrm{K}\) is released from Earth's surface to travel upward. If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give your answer in kilometers. [Hint: When the molecule comes to rest, the potential energy of the molecule will be \(m g h\) where \(m\) is the molecular mass in kilograms, \(g=9.81 \mathrm{m} \mathrm{s}^{-2}\) is the acceleration due to gravity, and \(h\) is the height, in meters, above Earth's surface.]

When solid \(\mathrm{KClO}_{3}\) is heated strongly, it decomposes to form solid potassium chloride, \(\mathrm{KCl}\), and \(\mathrm{O}_{2}\) gas. \(\mathrm{A}\) \(0.415 \mathrm{g}\) sample of impure \(\mathrm{KClO}_{3}\) is heated strongly and the \(\mathrm{O}_{2}\) gas produced by the decomposition is collected over water. When the wet \(\mathrm{O}_{2}\) gas is cooled back to \(26^{\circ} \mathrm{C}\), the total volume is \(229 \mathrm{mL}\) and the total pressure is 323 Torr. What is the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample? Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 25.22 Torr at \(26^{\circ} \mathrm{C}\).

A 2.35 L container of \(\mathrm{H}_{2}(\mathrm{g})\) at \(762 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\) is connected to a 3.17 L container of \(\mathrm{He}(\mathrm{g})\) at \(728 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\). After mixing, what is the total gas pressure, in millimeters of mercury, with the temperature remaining at \(24^{\circ} \mathrm{C} ?\)
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