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Chapter 5: Problem 13

Explain how Boyle鈥檚 law, Charles鈥檚 law, and Avogadro鈥檚 law are special cases of the ideal gas law

Short Answer

Expert verified
Boyle's Law, Charles's Law, and Avogadro's Law are special cases of the ideal gas law, with the equation \( PV = nRT \). Boyle's Law is a special case when temperature and the number of moles are constant, so \( P_1V_1 = P_2V_2 \). Charles's Law is a special case when pressure and the number of moles are constant, resulting in \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). Avogadro's Law is a special case when temperature and pressure are constant, giving \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \).

Step by step solution

01

Write down the ideal gas law equation

The ideal gas law equation is given by: \( PV = nRT \) where P is the pressure, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in Kelvin.
02

Analyze Boyle's Law

Boyle's Law states that for a fixed amount of an ideal gas at constant temperature (i.e., when n and T are held constant), the pressure and volume of the gas are inversely proportional. Mathematically, this can be expressed as: \( P_1V_1 = P_2V_2 \)
03

Show Boyle's Law as a special case of the ideal gas law

Since n and T are constant in Boyle's Law, we can rewrite the ideal gas law equation as: \( PV = constant \) Now, let's substitute the initial pressure and volume as \( P_1 \) and \( V_1 \) and final pressure and volume as \( P_2 \) and \( V_2 \) in the ideal gas law equation: \( P_1V_1 = nRT \) and \( P_2V_2 = nRT \) Since both equations are equal to \( nRT \), we can equate the left-hand sides of the two equations: \( P_1V_1 = P_2V_2 \) Thus, Boyle's Law is a special case of the ideal gas law when the temperature and the number of moles are held constant.
04

Analyze Charles's Law

Charles's Law states that for a fixed amount of an ideal gas at constant pressure (i.e., when n and P are held constant), the volume and temperature of the gas are directly proportional. Mathematically, this can be expressed as: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
05

Show Charles's Law as a special case of the ideal gas law

Since n and P are constant in Charles's Law, we can rewrite the ideal gas law equation as: \( V = \frac{nRT}{P} \) Now, let's substitute the initial volume and temperature as \( V_1 \) and \( T_1 \) and the final volume and temperature as \( V_2 \) and \( T_2 \) in the ideal gas law equation: \( V_1 = \frac{nRT_1}{P} \) and \( V_2 = \frac{nRT_2}{P} \) Dividing the first equation by the second equation gives: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) Thus, Charles's Law is a special case of the ideal gas law when the pressure and the number of moles are held constant.
06

Analyze Avogadro's Law

Avogadro's Law states that for a gas at constant temperature and pressure (i.e., when T and P are held constant), the volume of the gas is directly proportional to the number of moles. Mathematically, this can be expressed as: \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \)
07

Show Avogadro's Law as a special case of the ideal gas law

Since T and P are constant in Avogadro's Law, we can rewrite the ideal gas law equation as: \( V = \frac{nRT}{P} \) Now, let's substitute the initial volume and moles as \( V_1 \) and \( n_1 \) and the final volume and moles as \( V_2 \) and \( n_2 \) in the ideal gas law equation: \( V_1 = \frac{n_1RT}{P} \) and \( V_2 = \frac{n_2RT}{P} \) Dividing the first equation by the second equation gives: \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \) Thus, Avogadro's Law is a special case of the ideal gas law when the temperature and pressure are held constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boyle's Law
Boyle's Law beautifully explains how the pressure of a gas relates to its volume when the temperature and the number of gas moles stay the same. Imagine a balloon. When you squeeze it, the volume reduces, and the pressure inside increases. This is Boyle's Law in action, encapsulating the concept of inverse proportionality between pressure and volume. If gas is compressed into a smaller volume, its pressure increases proportionally. In the mathematical world, this is expressed as:
  • Initial state: \( P_1V_1 \)
  • Final state: \( P_2V_2 \)
This shows that \( P_1V_1 = P_2V_2 \) when temperature and the number of moles are constant, defining Boyle's Law as a special instance under the overarching ideal gas law.
Charles's Law
Charles's Law uncovers the relationship between the volume and temperature of a gas when pressure and amount of gas remain fixed. Picture a hot air balloon. As the air inside heats up, it expands. This demonstrates Charles's Law, where the volume of gas expands with increased temperature, provided the amount and pressure stay the same. Mathematically, this translates to:
  • Initial state: \( \frac{V_1}{T_1} \)
  • Final state: \( \frac{V_2}{T_2} \)
Therefore, it is concluded that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). This relationship highlights Charles's Law fitting snugly as a specific case of the ideal gas law when pressure and moles are constant.
Avogadro's Law
Avogadro's Law dives into how the amount of gas moles influences its volume at steady temperature and pressure. Consider inflating a tire. The more air you pump in, the larger the tire becomes. This is Avogadro's Law coming into play, where the volume increases with the number of gas moles, assuming constant temperature and pressure. Expressed mathematically, we have:
  • Initial state: \( \frac{V_1}{n_1} \)
  • Final state: \( \frac{V_2}{n_2} \)
It then stands that \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \). Thus, Avogadro's Law reveals itself as a salient facet of the ideal gas law when temperature and pressure are held stable.

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Most popular questions from this chapter

Consider separate \(2.5-\) L gaseous samples of \(\mathrm{He}\), \(\mathrm{N}_{2},\) and \(\mathrm{F}_{2},\) all at \(\mathrm{STP}\) and all acting ideally. Rank the gases in order of increasing average kinetic energy and in order of increasing average velocity.

You have a helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C} .\) You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0\(\%\) nitrogen and 21.0\(\%\) oxygen by volume. The 鈥渓ift鈥� of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C}\) . Assume atmospheric conditions are 1.00 atm and \(25^{\circ} \mathrm{C} .\)

A \(1.0-\mathrm{L}\) sample of air is collected at \(25^{\circ} \mathrm{C}\) at sea level \((1.00 \mathrm{atm}) .\) Estimate the volume this sample of air would have at an altitude of 15 \(\mathrm{km}\) ( see Fig.5.30) .At \(15 \mathrm{km},\) the pressure is about 0.1 \(\mathrm{atm} .\)

A steel cylinder contains 5.00 mole of graphite (pure carbon) and 5.00 moles of \(\mathrm{O}_{2} .\) The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0\(\%\) . Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A 150 -L reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C} :\) $$\begin{aligned} P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\ P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\ P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa} \end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)
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