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Suppose you have a fixed amount of an ideal gas at a constant volume. If the pressure of the gas is doubled while the volume is held constant, what happens to its temperature? [Section 10.4]

Short Answer

Expert verified
When the pressure of an ideal gas is doubled while the volume is held constant, the temperature of the gas also doubles: \(T2 = 2T1\).

Step by step solution

01

Write down the ideal gas law formula

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

Determine the relation between the initial and final states of the gas

Let the initial pressure be represented by P1 and the final pressure be represented by P2. Similarly, let the initial temperature be represented by T1 and the final temperature be represented by T2. According to the problem, the pressure is doubled, so: \(P2 = 2P1\)
03

Apply the ideal gas law for the initial and final states

Since the amount of gas enclosed in the container and the volume are constant, we can apply the ideal gas law for the initial state and the final state separately: Initial state: \(P1V = nRT1\) Final state: \(P2V = nRT2\)
04

Solve for the final temperature

We can rearrange the ideal gas law equations to express the temperatures as: \(T1 = \frac{P1V}{nR}\) \(T2 = \frac{P2V}{nR}\) Now, substitute \(P2 = 2P1\) into the equation for T2: \(T2 = \frac{2P1V}{nR}\) Now, from the equation for T1, we can see that \(2P1V = nR(2T1)\). Replacing this in the equation for T2, we get: \(T2 = 2T1\)
05

Conclude the relationship between the initial and final temperatures

The final temperature T2 is double the initial temperature T1: \(T2 = 2T1\) Therefore, when the pressure of an ideal gas is doubled while the volume is held constant, the temperature of the gas also doubles.

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Key Concepts

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

Pressure-Volume Relationship
Understanding the pressure-volume relationship in gases is essential to grasp the behavior of gases under various conditions. This relationship is commonly described by Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas remain constant.

In mathematical terms, Boyle's Law can be expressed as: \[ P \times V = \text{constant} \] This means that if the volume of a gas is decreased, the pressure increases, assuming no change in the gas's temperature or the amount (moles) of gas. Conversely, if the volume is increased, the pressure decreases. This is crucial in understanding how pressure changes when we manipulate the volume of a gas within a closed system. Boyle's Law is a stepping stone towards understanding more complex gas behaviors described by the ideal gas law.
Temperature Change in Gases
Temperature changes in gases are intimately linked with their pressure and volume changes, a concept incorporated in Charles's Law and Gay-Lussac's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature when the pressure is kept constant. Gay-Lussac's Law, on the other hand, states that the pressure of a gas is directly proportional to its temperature when the volume is kept constant.

These laws imply that if you were to heat a gas, expecting its volume to remain unchanged, the gas's pressure must increase to accommodate the increased kinetic energy of the particles. This is exactly the scenario presented in the textbook exercise above. By doubling the pressure while maintaining constant volume, you are indirectly deducing that the temperature must also have doubled to maintain the equilibrium described by the ideal gas law, as long as the amount of the gas remains unchanged.
Ideal Gas Constant
The ideal gas constant, denoted as 'R', is a fundamental parameter in the ideal gas law equation. It serves as a bridge linking pressure, volume, temperature, and the amount of a gas in moles. The value of R is the same for all gases, which is why it is termed a 'constant.'

Expressed in different units depending on the given pressure and volume units, the most common value used is approximately \(8.314 \text{J/(mol K)}\). The ideal gas constant helps to quantify the behavior of an ideal gas, allowing scientists and engineers to make predictions about the behavior of gases under different conditions. It is crucial to note that the ideal gas constant has different values depending on the units used for pressure, volume, and temperature. Ensuring that these units are consistent with those of 'R' is key to correctly using the ideal gas law and avoiding errors in calculations.

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Most popular questions from this chapter

Explain the difference between average speed and root-meansquare speed. Which is larger for a given gas sample at a fixed temperature?

In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for Xe, calculate the radius of a Xe atom and compare it to the value found in Figure 7.6, \(1.30 \AA\) A. Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

In the United States, barometric pressures are generally reported in inches of mercury (in. Hg). On a beautiful summer day in Chicago the barometric pressure is 30.45 in. Hg. (a) Convert this pressure to torr. (b) Convert this pressure to atm. (c) A meteorologist explains the nice weather by referring to a "high- pressure area." In light of your answer to parts (a) and (b), explain why this term makes sense.

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.452 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27{ }^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

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