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Chapter 19: Problem 4

A quantity of ideal gas at \(10.0^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) occupies a volume of \(2.50 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? (b) If the pressure is now raised to \(300 \mathrm{kPa}\) and the temperature is raised to \(30.0^{\circ} \mathrm{C}\), how much volume does the gas occupy? Assume no leaks.

Short Answer

Expert verified
(a) 10.70 moles; (b) 0.894 m鲁 volume.

Step by step solution

01

Convert Temperature to Kelvin

The temperature from Celsius to Kelvin can be converted using the formula: \[ T(K) = T(掳C) + 273.15 \] For the initial temperature:\[ T_{1} = 10.0 + 273.15 = 283.15 \, \text{K} \]For the final temperature:\[ T_{2} = 30.0 + 273.15 = 303.15 \, \text{K} \]
02

Use Ideal Gas Law for Initial State

The ideal gas law relates pressure, volume, temperature, and moles. It's given as:\[ PV = nRT \]Where:- \( P \) is pressure in pascals \((100 \text{ kPa} = 100,000 \text{ Pa})\)- \( V \) is volume in cubic meters \((2.50 \text{ m}^3)\)- \( n \) is the number of moles- \( R \) is the ideal gas constant \((8.314 \, \text{J/(mol K)})\)- \( T \) is temperature in Kelvin \((283.15 \, \text{K})\)Rearranging for \( n \):\[ n = \frac{PV}{RT} = \frac{100,000 \times 2.50}{8.314 \times 283.15} \]\[ n \approx 10.70 \, \text{moles} \]
03

Apply the Combined Gas Law for Final State

The final state of gas can be found using the combined gas law assuming constant moles:\[ \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}} \]Solving for \( V_{2} \):\[ V_{2} = \frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}} = \frac{100,000 \times 2.50 \times 303.15}{300,000 \times 283.15} \]\[ V_{2} \approx 0.894 \, \text{m}^3 \]
04

Finalize Answers

From the calculations:- (a) The number of moles is approximately \( 10.70 \) moles.- (b) Upon increasing pressure to \( 300 \text{ kPa} \) and temperature to \( 30.0^{\circ} \text{C} \), the gas occupies a volume of approximately \( 0.894 \, \text{m}^3 \).

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Key Concepts

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

Combined Gas Law
The Combined Gas Law is a convenient tool used in chemistry to explore how a gas behaves when temperature, pressure, and volume change. It's a combination of Boyle's Law, Charles's Law, and Gay-Lussac's Law. This law helps predict one state property when the other three are altered, **without the number of moles changing**. The formula is: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]- **\(P_1, V_1, T_1\):** Initial pressure, volume, and temperature.- **\(P_2, V_2, T_2\):** Final pressure, volume, and temperature.This law is crucial in situations like Earth鈥檚 atmospheric changes, diving, and when dealing with gases in industrial processes. It elegantly ties with the Ideal Gas Law to account for real-world gas transformations. When using it, remember to keep temperature in Kelvin and pressure in the same units throughout your calculations.
This ensures consistency and accuracy in your results.In our example, knowing initial conditions and how they changed allows us to find the final volume of the gas.
moles
In chemistry, a mole is a fundamental unit for measuring amount of substance. It allows chemists to count particles such as atoms, molecules, or ions in a given sample. One mole is equivalent to Avogadro鈥檚 number, which is approximately **6.022 x 10虏鲁 particles**.In the context of gases, moles tie into gas laws via the Ideal Gas Law: \[PV = nRT\] where \(n\) indicates the number of moles.**Why are moles so important?**- Understanding reaction stoichiometry.- Converting between mass and number of particles.In our problem, by using the Ideal Gas Law, we calculate the moles of gas based on provided pressure, volume, and temperature. This value, around **10.70 moles**, serves as a cornerstone for further calculations using the Combined Gas Law.
volume calculation
Determining the volume a gas occupies under different conditions is a critical application in understanding gas behavior. For an ideal gas, volume relates directly to pressure and temperature through laws such as the Ideal Gas Law and Combined Gas Law. In our case, we start by finding the initial number of moles using the Ideal Gas Law. Then, by applying the Combined Gas Law, we determine the new volume after changes in pressure and temperature. Steps involved: - Calculate moles using the initial conditions. - Use the Combined Gas Law, with the kept constant, to find the new volume. In the exercise example, upon increasing the pressure to **300 kPa** and temperature to **30掳C**, calculations show a final volume of about **0.894 m鲁**. Always double-check the unit consistency for accurate results, notably ensuring temperature is in Kelvin.
  • Initial conditions gave us the key moles value.
  • New conditions show us how the volume shifts.
temperature conversion
Temperature conversion, especially to Kelvin, is essential when dealing with gas laws. Gases are sensitive to temperature changes, affecting their pressure and volume. The Kelvin scale is preferred because it's an absolute scale, meaning there are no negative values. To convert from Celsius to Kelvin, use:\[T(K) = T(掳C) + 273.15\]**Example**:For a temperature of **10掳C**:\[T(K) = 10 + 273.15 = 283.15 \, \text{K}\]And, **30掳C**:\[T(K) = 30 + 273.15 = 303.15 \, \text{K}\]Accurate temperature conversion is vital for calculations since gas laws, like the Ideal Gas Law and Combined Gas Law, require temperatures in Kelvin.
  • Ensures consistency across variables.
  • Allows for straightforward comparison and calculations.
Be sure to use Kelvin in all formulas to prevent errors in outputs and to gain an accurate understanding of gas behavior under varying conditions.

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Most popular questions from this chapter

The volume of an ideal gas is adiabatically reduced from \(200 \mathrm{~L}\) to \(74.3 \mathrm{~L}\). The initial pressure and temperature are \(1.00 \mathrm{~atm}\) and \(300 \mathrm{~K}\). The final pressure is 4.00 atm. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is the final temperature? (c) How many moles are in the gas?

Suppose \(1.80 \mathrm{~mol}\) of an ideal gas is taken from a volume of \(3.00 \mathrm{~m}^{3}\) to a volume of \(1.50 \mathrm{~m}^{3}\) via an isothermal compression at \(30^{\circ} \mathrm{C}\). (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?

An ideal gas, at initial temperature \(T_{1}\) and initial volume \(2.0 \mathrm{~m}^{3},\) is expanded adiabatically to a volume of \(4.0 \mathrm{~m}^{3},\) then expanded isothermally to a volume of \(10 \mathrm{~m}^{3},\) and then compressed adiabatically back to \(T_{1} .\) What is its final volume?

Adiabatic wind. The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure \(p\) depends on altitude \(y\) according to \(p=p_{0} \exp (-a y),\) where \(p_{0}=1.00 \mathrm{~atm}\) and \(a=1.16 \times 10^{-4} \mathrm{~m}^{-1} .\) Also assume that the ratio of the molar specific heats is \(\gamma=\frac{4}{3}\). A parcel of air with an initial temperature of \(-5.00^{\circ} \mathrm{C}\) descends adiabatically from \(y_{1}=4267 \mathrm{~m}\) to \(y=1567 \mathrm{~m} .\) What is its temperature at the end of the descent?

A beam of hydrogen molecules \(\left(\mathrm{H}_{2}\right)\) is directed toward a wall, at an angle of \(55^{\circ}\) with the normal to the wall. Each molecule in the beam has a speed of \(1.0 \mathrm{~km} / \mathrm{s}\) and a mass of \(3.3 \times 10^{-24} \mathrm{~g}\). The beam strikes the wall over an area of \(2.0 \mathrm{~cm}^{2},\) at the rate of \(10^{23} \mathrm{~mol}-\) ecules per second. What is the beam's pressure on the wall?
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