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You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant. ll be produced?

Short Answer

Expert verified
Pushing down on the piston to halve the volume while keeping the temperature constant (action c) would double the gas pressure.

Step by step solution

01

Action (a): Doubling the volume while keeping the temperature constant.

For this action, the temperature and the number of moles of gas remain constant. We only need to consider the volume (V) and the pressure (P): \(PV = \text{constant}\). If the volume is doubled and the product PV remains constant, the pressure will be halved (P/2), and not doubled. So action (a) will not double the gas pressure.
02

Action (b): Doubling the temperature while keeping the volume constant.

For this action, the volume and the number of moles of gas remain constant. We only need to consider the temperature (T) and the pressure (P): \(\frac{P}{T} = \text{constant}\). When the temperature is increased from 25°C to 50°C, we must first convert it to Kelvin: 25°C + 273.15 = 298.15K and 50°C + 273.15 = 323.15K. To check the effect on the pressure, let's write the relation for the initial and final states: \[ \frac{P_i}{T_i} = \frac{P_f}{T_f}. \] With the given temperature values: \[ \frac{P_i}{298.15} = \frac{P_f}{323.15}. \] Solving for the final pressure (\(P_f\)): \[ P_f = P_i \times \frac{323.15}{298.15}. \] Since \(323.15 / 298.15 \approx 1.084\), the pressure will not be doubled (it would be increased by about 8.4% only). So action (b) will not double the gas pressure.
03

Action (c): Halving the volume while keeping the temperature constant.

For this action, the temperature and the number of moles of gas remain constant. We only need to consider the volume (V) and the pressure (P): \(PV = \text{constant}\). If the volume is halved and the product PV remains constant, the pressure will be doubled (2P), which is our objective. So action (c) will double the gas pressure. To sum up the results: 1. Action (a) will not double the gas pressure. 2. Action (b) will not double the gas pressure. 3. Action (c) will double the gas pressure. Therefore, pushing down on the piston to halve the volume while keeping the temperature constant (action c) would double the gas pressure.

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Key Concepts

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

Boyle's Law
Understanding gas pressure in chemistry often begins with Boyle's Law, a principle which deals with the inversely proportional relationship between the pressure and volume of a gas at constant temperature. This means that if you increase the volume of gas in a container, its pressure decreases, and conversely, if you decrease the volume, the pressure increases.

From our exercise, when pushing down on a piston to halve the volume (action c), Boyle’s Law can be applied to predict the outcome on the pressure of the gas. Mathematically, Boyle’s Law is represented as \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively. Halving the volume while maintaining the same amount of gas and constant temperature will therefore double the pressure, as we've seen in the solution to action (c).
Charles's Law
Another fundamental aspect of gas behavior can be captured by Charles's Law, which states that the volume of a gas is directly proportional to its absolute temperature, assuming the number of gas particles and the pressure remain constant.

Applying Charles's Law to action (b) from our exercise, where the temperature increases from \(25^\circ \mathrm{C}\) to \(50^\circ \mathrm{C}\) while the volume remains constant, results in an increase in pressure. This is due to the direct relationship between temperature and pressure when volume does not change. Charles's Law is mathematically expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( T \) must be in Kelvin. In practice, though, action (b) doesn't double the pressure, as the temperature increase to \( 50^\circ \mathrm{C} \) results in a rise in pressure of only about 8.4%, not the 100% needed to double it.
Ideal Gas Law
The Ideal Gas Law is a more encompassing equation that relates pressure (P), volume (V), temperature (T), and the number of moles of gas (n) to describe the state of a hypothetical 'ideal' gas. The law is commonly expressed as \( PV = nRT \), where R is the ideal gas constant. This equation meshes the principles of Boyle's Law, Charles's Law, and others into one formula.

With respect to our exercise, the Ideal Gas Law can be used to predict the effect of changing more than one condition of a gas at a time. However, since in our step-by-step solution we're changing one condition at a time—volume in action (a) and (c), and temperature in action (b)—the individual laws are sufficient for prediction.
Temperature and Pressure Relationship
The temperature and pressure relationship of gases is another fundamental concept to grasp when studying gas law in chemistry. According to Gay-Lussac's Law, which is a component of the Ideal Gas Law, the pressure of a gas is directly proportional to its temperature when the volume is kept constant.

In our text exercise, action (b) attempts to double the pressure by increasing temperature while keeping volume constant. This is an application of Gay-Lussac's Law, and while pressure does increase with temperature, the specific increase from \(25^\circ \mathrm{C}\) to \(50^\circ \mathrm{C}\) is not sufficient to double the pressure. This helps illustrate the significance of understanding this temperature-pressure relationship to correctly predict the behavior of gases under various conditions.

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Most popular questions from this chapter

Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(\mathrm{m}^{1 / 2}\). (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\). Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and 1 atm pressure. (d) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in L-bar/mol-K.

To derive the ideal-gas equation, we assume that the volume of the gas atoms/molecules can be neglected. Given the atomic radius of neon, \(0.69 \AA\), and knowing that a sphere has a volume of \(4 \pi \mathrm{r}^{3} / 3\), calculate the fraction of space that Ne atoms occupy in a sample of neon at STP.

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$ \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22^{\circ} \mathrm{C}\) ?

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gaseous state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.

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