Question

Suppose 0.03127 mole of an ideal monatomic gas with a pressure of 1.000 atm at \(273.15 \mathrm{~K}\) is cooled at constant pressure and reduced in volume. The amount of work done by the gas is \(-3.404 \cdot 10^{1} \mathrm{~J}\). By what percentage was the volume of the gas reduced?

Short answer

Based on the given information and calculations, the volume of the ideal monatomic gas increased by approximately 1.472% rather than reduced after cooling at constant pressure.

Step-by-step solution

Identify the given values

We are given: - moles of gas (n) = 0.03127 - initial temperature (T1) = 273.15 K - initial pressure (P1) = 1.000 atm - work done by the gas (W) = -3.404 * 10^1 J Note that the pressure is constant throughout the process.

Convert pressure to SI units

Convert the pressure in atm to pascals (Pa): 1 atm = 101325 Pa P1 = 1.000 * 101325 = 101325 Pa

Calculate the initial volume using the ideal gas law

The ideal gas law is given by \(PV = nRT\), where R is the ideal gas constant. In SI units, R = 8.314 J/(mol * K). Calculate the initial volume (V1) with the given values: V1 = (n * R * T1) / P1 V1 = (0.03127 * 8.314 * 273.15) / 101325 V1 ≈ 0.02287 m^3

Calculate the final temperature using work done (isobaric process)

In an isobaric process, the work done by the gas is given by \(W = -P \Delta V\), and \(\Delta V = V2 - V1\). First, calculate \(\Delta V\) with the work done and pressure: \(\Delta V = -\frac{W}{P}\) \(\Delta V = -\frac{-3.404 * 10^1}{101325}\) \(\Delta V ≈ 3.361 * 10^{-4} \mathrm{~m^3}\) Now, calculate the final volume (V2): V2 = V1 + \(\Delta V\) V2 ≈ 0.02287 + 3.361 * 10^{-4} V2 ≈ 0.023206 m^3

Calculate the final temperature using the ideal gas law

Rearrange the ideal gas law to solve for the final temperature (T2): T2 = (P * V2) / (n * R) T2 = (101325 * 0.023206) / (0.03127 * 8.314) T2 ≈ 96.82 K

Calculate the percentage reduction in volume

Find the difference between the initial and final volumes, and calculate the percentage change in volume: Percentage reduction in volume = \(\frac{(V1 - V2)}{V1} * 100\) Percentage reduction in volume = \(\frac{(0.02287 - 0.023206)}{0.02287} * 100\) Percentage reduction in volume ≈ -1.472 % Here, the negative sign indicates that the volume has actually increased, not reduced. Therefore, the volume of the gas was increased by about 1.472%.

Key concepts

Isobaric Process

An isobaric process is a thermodynamic process in which the pressure remains constant. This constraint simplifies the relations between various thermodynamic properties. For an ideal gas, the ideal gas law, which is expressed as \(PV = nRT\), where P stands for pressure, V for volume, n for the number of moles, R for the ideal gas constant, and T for the temperature in Kelvin, is applicable. During an isobaric process, as the temperature of the gas changes, the volume must also change to maintain constant pressure. The process is governed by the relationship \(W = -P\triangle V\), characterizing work done by the gas. In the context of our exercise, the gas is initially at a known temperature and volume, and it undergoes a reduction in temperature. Since the pressure is constant, the reduction in temperature implies a change in volume, which in turn, allows us to calculate the work done by the gas during this process. Understanding isobaric processes is important because it helps us to predict how gases will behave when they are heated or cooled, such as in heating systems, car engines, or even the earth's atmosphere.

Work Done by Gas

The work done by a gas during a thermodynamic process is an essential concept in physics and engineering. It is the energy transferred by the gas to its surroundings, typically resulting from a volume change under constant pressure. In the case of an isobaric process, the formula \(W = -P\triangle V\) is used, with the negative sign indicating that work is done by the gas when the volume decreases and implying work is done on the gas when the volume increases. For our specific exercise, calculating the work done involves using the value of pressure, which must first be converted to the correct units (Pa), and the change in volume deduced from the exercise information. The 'negative' work here suggests that the gas is indeed doing work on the surroundings. It's essential to have a strong grasp of this concept as it applies to everyday phenomena such as inflating a balloon or the expansion of pistons in a car engine.

Volume Reduction Percentage

Volume reduction percentage is a measure of how much a gas's volume decreases relative to its original volume during a particular process, expressed as a percentage. Calculating this percentage provides insight into the extent of the volume change. The formula to determine the volume reduction percentage is \(\frac{(V1 - V2)}{V1} \times 100\), where V1 is the initial volume and V2 is the final volume. In our exercise, once we have found the values for the initial and final volumes, we subtract V2 from V1, then divide by V1, and finally multiply by 100 to get the percentage. Contrary to the assumption that the volume would decrease (denoted by a positive percentage), our calculation revealed a 'negative' volume reduction percentage, indicating that the volume has increased, a crucial point to note as volume reduction percentage can signify an expansion rather than a reduction. Such calculations are not just academic; they apply in various contexts, such as understanding gas compressibility in engine cylinders, or even the expansion of air in weather balloons.