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Chapter 5: Problem 19

A gas occupying a volume of \(725 \mathrm{~mL}\) at a pressure of 0.970 atm is allowed to expand at constant temperature until its pressure reaches 0.541 atm. What is its final volume?

Short Answer

Expert verified
After performing the calculation, the final volume \(V_2\) comes out to be approximately \(1378 \mathrm{~mL}\).

Step by step solution

01

Understand and write down the given information

Initially, the gas has a volume of \(725 \mathrm{~mL}\) or \(V_1 = 725 \mathrm{~mL}\) and pressure of \(0.970 \mathrm{~atm}\) or \(P_1 = 0.970 \mathrm{~atm}\). Finally, the pressure of the gas is \(0.541 \mathrm{~atm}\) or \(P_2 = 0.541 \mathrm{~atm}\). The goal is to find the final volume \(V_2\).
02

Apply Boyle's Law

Boyle's Law states that the product of initial pressure \(P_1\) and initial volume \(V_1\) equals the product of final pressure \(P_2\) and final volume \(V_2\) given that temperature remains constant. This can be written as \(P_1V_1 = P_2V_2\).
03

Calculate the final volume

Rearrange the equation to solve for the final volume \(V_2\): \(V_2 = \frac{P_1V_1}{P_2}\). Substituting the given values, \(V_2 = \frac{0.970 \mathrm{~atm} \times 725 \mathrm{~mL}}{0.541 \mathrm{~atm}}\).

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Most popular questions from this chapter

The following apparatus can be used to measure atomic and molecular speed. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm}\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (m/s) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r,\) where \(r\) is the radius. \()\) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the Bi atoms. Compare your result in (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

A student breaks a thermometer and spills most of the mercury (Hg) onto the floor of a laboratory that measures \(15.2 \mathrm{~m}\) long, \(6.6 \mathrm{~m}\) wide, and \(2.4 \mathrm{~m}\) high. (a) Calculate the mass of mercury vapor (in grams) in the room at \(20^{\circ} \mathrm{C}\). The vapor pressure of mercury at \(20^{\circ} \mathrm{C}\) is \(1.7 \times 10^{-6} \mathrm{~atm} .\) (b) Does the concentration of mercury vapor exceed the air quality regulation of \(0.050 \mathrm{mg} \mathrm{Hg} / \mathrm{m}^{3}\) of air? (c) One way to treat small quantities of spilled mercury is to spray sulfur powder over the metal. Suggest a physical and a chemical reason for this action.

A mixture of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{MgCO}_{3}\) of mass \(7.63 \mathrm{~g}\) is reacted with an excess of hydrochloric acid. The \(\mathrm{CO}_{2}\) gas generated occupies a volume of \(1.67 \mathrm{~L}\) at 1.24 atm and \(26^{\circ} \mathrm{C}\). From these data, calculate the percent composition by mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in the mixture.

(a) A real gas is introduced into a flask of volume \(V\). Is the corrected volume of the gas greater or less than \(V ?\) (b) Ammonia has a larger \(a\) value than neon does (see Table 5.4 ). What can you conclude about the relative strength of the attractive forces between molecules of ammonia and between atoms of neon?

(a) What volumes (in liters) of ammonia and oxygen must react to form \(12.8 \mathrm{~L}\) of nitric oxide according to the equation at the same temperature and pressure? $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ (b) What volumes (in liters) of propane and water vapor must react to form \(8.96 \mathrm{~L}\) of hydrogen according to the equation at the same temperature and pressure? $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{CO}(g)+7 \mathrm{H}_{2}(g) $$
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