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Chapter 8: Problem 150

Which of the following statements is(are) true? a. If the number of moles of a gas is doubled, the volume will double, assuming the pressure and temperature of the gas remain constant. b. If the temperature of a gas increases from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) the volume of the gas would double, assuming that the pressure and the number of moles of gas remain constant. c. The device that measures atmospheric pressure is called a barometer. d. If the volume of a gas decreases by one half, then the pressure would double, assuming that the number of moles and the temperature of the gas remain constant.

Short Answer

Expert verified
The true statements are: a, c, and d.

Step by step solution

01

Statement a: Doubling the number of moles, volume will double

Using the Ideal Gas Law, \(PV = nRT\), assume the pressure and temperature are constant. If we double the number of moles, we now have \(P(2V) = 2nRT\). Divide this equation by the original ideal gas equation, \(PV = nRT\), to get \(\frac{2V}{V} = \frac{2nRT}{nRT}\), which simplifies to 2 = 2. This shows that statement a is true.
02

Statement b: Doubling the temperature, volume will double

Again, using the Ideal Gas Law, \(PV = nRT\), assume the pressure and the number of moles are constant. Convert the given temperatures from Celsius to Kelvin by adding 273.15. So, we have that T1 = 273.15 K + 25 K = 298.15 K, and T2 = 273.15 K + 50 K = 323.15 K. Now let's calculate the ratio of the volumes with the given temperatures: \(\frac{V_2}{V_1} = \frac{nRT_2}{nRT_1}\). The number of moles and the gas constant R are constant, so we get \(\frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{323.15 K}{298.15 K} \approx 1.08\). Since the volume ratio is not equal to 2, statement b is false.
03

Statement c: Barometer measures atmospheric pressure

A barometer is an instrument that is used to measure atmospheric pressure. This statement is just a definition and is true.
04

Statement d: Halving the volume, pressure will double

From the Ideal Gas Law, \(PV = nRT\), let's assume the number of moles and the temperature are constant. If the volume decreases by half, then the equation becomes \(P \times \frac{V}{2} = nRT\). We can rearrange this equation to find the relationship between the original pressure, P, and the new pressure, P': \( P' = \frac{2nRT}{V} = 2P\). This shows that statement d is true as well. In conclusion, the true statements are: a, c, and d.

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Key Concepts

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

Gas Behavior
Gas behavior can be understood through the lens of the Ideal Gas Law, represented by the equation \(PV = nRT\). This equation incorporates several key factors:
  • \(P\) is the pressure exerted by the gas.
  • \(V\) is the volume occupied by the gas.
  • \(n\) signifies the number of moles of the gas.
  • \(R\) is the ideal gas constant, providing a bridge between the other units.
  • \(T\) represents the temperature in Kelvin.
To illustrate, if you were to double the number of moles \(n\), keeping both temperature \(T\) and pressure \(P\) constant, the volume \(V\) would inevitably double to maintain the equilibrium indicated by the Ideal Gas Law. This constant interdependence shows how gases expand or compress in response to changes in volume, temperature, or the amount of gas. Understanding these behaviors allows us to predict how gases will react under different conditions.
Temperature and Volume Relationship
The relationship between temperature and volume in a gas highlights a fundamental principle known as Charles's Law. It articulates that the volume of a gas is directly proportional to its temperature, provided the pressure and the moles of gas are kept constant. Let's break this down:
  • In our exercise, temperatures in Celsius must be first converted to Kelvin to apply the Ideal Gas Law accurately: \(T_{Kelvin} = T_{Celsius} + 273.15\).
  • Increasing the temperature from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) translates from \(298.15\, \mathrm{K}\) to \(323.15\, \mathrm{K}\) in Kelvin.
  • The ratio \(\frac{V_2}{V_1} = \frac{T_2}{T_1}\) implies a dependence of volume on temperature.
With temperatures \(T_1\) and \(T_2\), the actual volume ratio is \(1.08\), showing that the volume does not quite double when temperature increases by \(25^{\circ} \mathrm{C}\). This reflects why statement b was labeled false. A thorough comprehension of this relationship is crucial in real-life applications such as engineering, where precise temperature controls are essential.
Atmospheric Pressure Measurement
Measuring atmospheric pressure is an essential aspect of understanding weather patterns and predicting climate changes. A barometer is the standard instrument used for this purpose.
  • Barometers usually contain mercury or aneroid sensors to gauge the pressure exerted by the earth's atmosphere.
  • The height of mercury in the tube is directly proportional to atmospheric pressure鈥攁 principle dating back to Torricelli's time.
  • This pressure measurement can infer weather changes; for instance, falling pressure typically indicates stormy weather.
In our exercise, the definition provided in statement c regarding the function of a barometer was correct, hence labeled as true. By understanding atmospheric pressure, we become better equipped to handle environmental challenges and navigate daily activities impacted by the weather.

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Most popular questions from this chapter

A sealed balloon is filled with \(1.00 \mathrm{L}\) helium at \(23^{\circ} \mathrm{C}\) and 1.00 atm. The balloon rises to a point in the atmosphere where the pressure is \(220 .\) torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of \(220 .\) torr?

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(s)\) : $$\text { Compound } \longrightarrow\mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ The product gas is then passed through a concentrated solution of KOH to remove the \(\mathrm{CO}_{2}\). After passage through the KOH solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a \(0.253-g\) sample of a compound produced \(31.8 \mathrm{mL} \mathrm{N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and \(726\) torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is 23.8 torr.)

Complete the following table for an ideal gas. $$\begin{array}{|lccc|}\hline P(\mathrm{atm}) & V(\mathrm{L}) & n(\mathrm{mol}) & T \\ \hline 5.00 & & 2.00 & 155^{\circ} \mathrm{C} \\\\\hline0.300 & 2.00 & & 155 \mathrm{K} \\ \hline 4.47 & 25.0 & 2.01 & \\\\\hline & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\\ \hline\end{array}$$

Draw a qualitative graph to show how the first property varies with the second in each of the following (assume 1 mole of an ideal gas and \(T\) in kelvin). a. \(P V\) versus \(V\) with constant \(T\) b. \(P\) versus \(T\) with constant \(V\) c. \(T\) versus \(V\) with constant \(P\) d. \(P\) versus \(V\) with constant \(T\) e. \(P\) versus \(1 / V\) with constant \(T\) f. \(P V / T\) versus \(P\)

Consider separate \(1.0\) -\(\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g),\) both at \(1.00\) atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?
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