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Chapter 5: Problem 21

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure changes from 760 torr to \(202 \mathrm{kPa}\), and the temperature changes from \(37^{\circ} \mathrm{C}\) to \(155 \mathrm{~K}\) (b) The temperature changes from \(305 \mathrm{~K}\) to \(32^{\circ} \mathrm{C},\) and the pressure changes from 2 atm to \(101 \mathrm{kPa}\). (c) The pressure is reduced by a factor of 4 (at constant \(T\) ).

Short Answer

Expert verified
a) Volume decreases to 6.38 L from 25.47 L.b) Volume increases to 24.98 L from 12.52 L.c) Volume increases by a factor of 4.

Step by step solution

01

- Understand the Ideal Gas Law

The Ideal Gas Law is given by the equation \[PV = nRT\], where: - \(P\) is the pressure in atmospheres, - \(V\) is the volume in liters, - \(n\) is the number of moles, - \(R\) is the universal gas constant (0.0821 atm路L/mol路K), - \(T\) is the temperature in Kelvin.
02

- Convert Given Values to Consistent Units (Part a)

Convert the given pressures and temperatures to consistent units (atm and K): \[202 \text{ kPa} = 202 \times 0.00986923 \text{ atm} = 1.993 \text{ atm} \] \[37^{\text{ o}} \text{C} = 37 + 273.15 = 310.15 \text{ K} \] \[155 \text{ K} (No conversion needed) \]
03

- Apply Ideal Gas Law (Part a)

Using the Ideal Gas Law equation for both initial and final states: Initial: \(P_1V_1 = nRT_1 \rightarrow 1 \text{ atm} \times V_1 = 1 \text{ mol} \times 0.0821 \text{ atm路L/mol路K} \times 310.15 \text{ K}\) \[ V_1 = 25.47 \text{ L} \] Final: \(P_2V_2 = nRT_2 \rightarrow 1.993 \text{ atm} \times V_2 = 1 \text{ mol} \times 0.0821 \text{ atm路L/mol路K} \times 155 \text{ K}\) \[ V_2 = 6.38 \text{ L} \]
04

- Convert Given Values to Consistent Units (Part b)

Convert the given pressures and temperatures to consistent units (atm and K): \[32^{\text{ o}} \text{C} = 32 + 273.15 = 305.15 \text{ K}\] \[101 \text{ kPa} = 101 \times 0.00986923 \text{ atm} = 0.998 \text{ atm}\] The initial temperature and pressure are already in consistent units: 305 K and 2 atm.
05

- Apply Ideal Gas Law (Part b)

Using the Ideal Gas Law equation for both initial and final states: Initial: \(P_1V_1 = nRT_1 \rightarrow 2 \text{ atm} \times V_1 = 1 \text{ mol} \times 0.0821 \text{ atm路L/mol路K} \times 305 \text{ K}\) \[ V_1 = 12.52 \text{ L} \] Final: \(P_2V_2 = nRT_2 \rightarrow 0.998 \text{ atm} \times V_2 = 1 \text{ mol} \times 0.0821 \text{ atm路L/mol路K} \times 305.15 \text{ K}\) \[ V_2 = 24.98 \text{ L} \]
06

- Apply the Pressure Reduction Factor (Part c)

For this scenario, the pressure is reduced by a factor of 4 while the temperature is held constant: Initially: \(PV = nRT\) means that \(P_1V = nRT \). When the pressure is reduced by a factor of 4: \(\frac{P_1}{4} \). According to the Ideal Gas Law, keeping \(T\) constant: \(\frac{P_1}{4} \times V_2 = nRT \). Since \(nRT = P_1V\): \[P_1V = \frac{P_1}{4} \times V_2 \rightarrow 1V = \frac{V_2}{4} \times 4\rightarrow V_2 = 4V_1 \]. Thus, the volume would increase by a factor of 4.

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Key Concepts

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

Pressure Conversion
In the context of the Ideal Gas Law, pressure is often measured in different units such as atmospheres (atm), kilopascals (kPa), or torr. To successfully apply the Ideal Gas Law, all pressures must be in the same units.

For example, converting from kPa to atm requires multiplying by 0.00986923, as seen in the conversion: 202 kPa = 202 脳 0.00986923 atm = 1.993 atm.

Converting from units like torr to atm involves dividing by 760, since 1 atm equals 760 torr. Ensuring consistent pressure units is essential for accurate calculations using the Ideal Gas Law.
Temperature Conversion
Temperature must be in Kelvin (K) when applying gas laws, as the Kelvin scale starts at absolute zero.

Converting from Celsius to Kelvin is straightforward: add 273.15 to the Celsius temperature. For instance, converting 37掳颁 to Kelvin results in: 37 + 273.15 = 310.15 K.

Conversely, converting back to Celsius involves subtracting 273.15 from the Kelvin temperature. Proper temperature conversion is vital for the Ideal Gas Law to work accurately, ensuring the correct relationship between pressure, volume, and temperature.
Gas Constant
The universal gas constant, denoted as R, plays a critical role in the Ideal Gas Law. Its value is 0.0821 atm路L/mol路K. This constant allows the bridging of units like atmospheres, liters, moles, and Kelvin in the equation PV = nRT.

R ensures that the relationship between pressure, volume, and temperature holds true for one mole of an ideal gas. Regardless of the state of the gas concerning pressure and temperature, applying R consistently ensures that calculations yield correct and meaningful results.
Effect on Volume
Understanding the effect of changes in pressure and temperature on the volume of a gas is crucial when using the Ideal Gas Law. According to the equation PV = nRT, volume varies directly with temperature and inversely with pressure.

For instance, if pressure is reduced by a factor of 4 (keeping temperature constant), the volume increases by a factor of 4. This is shown mathematically as:

Initially: PV = nRT

Reduced Pressure: P/4 * V2 = nRT. Since nRT = P * V1, it follows that V2 = 4 * V1.

Similarly, changes in temperature like from 310.15 K to 155 K, with other factors constant, would lead to a decrease in volume, emphasizing the direct relationship between temperature and volume.

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Most popular questions from this chapter

In a certain experiment, magnesium boride \(\left(\mathrm{Mg}_{3} \mathrm{~B}_{2}\right)\) reacted with acid to form a mixture of four boron hydrides \(\left(\mathrm{B}_{x} \mathrm{H}_{y}\right),\) three as liquids (labeled I, II, and III) and one as a gas (IV). (a) When a \(0.1000-\mathrm{g}\) sample of each liquid was transferred to an evacuated \(750.0-\mathrm{mL}\) container and volatilized at \(70.00^{\circ} \mathrm{C},\) sample I had a pressure of 0.05951 atm; sample II, 0.07045 atm; and sample III, 0.05767 atm. What is the molar mass of each liquid? (b) Boron was determined to be \(85.63 \%\) by mass in sample I, \(81.10 \%\) in II, and \(82.98 \%\) in III. What is the molecular formula of each sample? (c) Sample IV was found to be \(78.14 \%\) boron. The rate of effusion for this gas was compared to that of sulfur dioxide; under identical conditions, \(350.0 \mathrm{~mL}\) of sample IV effused in \(12.00 \mathrm{~min}\) and \(250.0 \mathrm{~mL}\) of sulfur dioxide effused in 13.04 min. What is the molecular formula of sample IV?

After \(0.600 \mathrm{~L}\) of \(\mathrm{Ar}\) at \(1.20 \mathrm{~atm}\) and \(227^{\circ} \mathrm{C}\) is mixed with \(0.200 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at 501 torr and \(127^{\circ} \mathrm{C}\) in a \(400-\mathrm{mL}\) flask at \(27^{\circ} \mathrm{C}\) what is the pressure in the flask?

Dilute aqueous hydrogen peroxide is used as a bleaching agent and for disinfecting surfaces and small cuts. Its concentration is sometimes given as a certain number of "volumes hydrogen peroxide," which refers to the number of volumes of \(\mathrm{O}_{2}\) gas, measured at STP, that a given volume of hydrogen peroxide solution will release when it decomposes to \(\mathrm{O}_{2}\) and liquid \(\mathrm{H}_{2} \mathrm{O}\). How many grams of hydrogen peroxide are in \(0.100 \mathrm{~L}\) of \(" 20\) volumes hydrogen peroxide" solution?

In preparation for a combustion demonstration, a professor fills a balloon with equal molar amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2},\) but the demonstration has to be postponed until the next day. During the night, both gases leak through pores in the balloon. If \(35 \%\) of the \(\mathrm{H}_{2}\) leaks, what is the \(\mathrm{O}_{2} / \mathrm{H}_{2}\) ratio in the balloon the next day?

An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a 500.-mL sample at 700 . torr and \(38^{\circ} \mathrm{C},\) she adds \(20.00 \mathrm{~mL}\) of \(0.01017 M\) aqueous iodine, which reacts as follows: \(\mathrm{SO}_{2}(g)+\mathrm{I}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$\mathrm{HSO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q)+\mathrm{H}^{+}(a q) \quad[\text { unbalanced }]$$ Excess \(\mathrm{I}_{2}\) reacts with \(11.37 \mathrm{~mL}\) of \(0.0105 \mathrm{M}\) sodium thiosulfate: \(\mathrm{I}_{2}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) \quad[\) unbalanced \(]\) What is the volume \(\%\) of \(\mathrm{SO}_{2}\) in the air sample?
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