Question

A spherical glass container of unknown volume contains helium gas at \(25^{\circ} \mathrm{C}\) and \(1.960\) atm. When a portion of the helium is withdrawn and adjusted to 1.00 atm at \(25^{\circ} \mathrm{C},\) it is found to have a volume of \(1.75 \mathrm{cm}^{3} .\) The gas remaining in the first container shows a pressure of \(1.710 \) atm. Calculate the volume of the spherical container.

Short answer

The volume of the spherical container is \(0.02185 \mathrm{L}\).

Step-by-step solution

Identify given values

From the exercise, we are given the following values: - Initial temperature: \(25^{\circ} \mathrm{C}\) (constant throughout the problem, so convert it to kelvin: \(T = 25 + 273.15 = 298.15 \mathrm{K}\)) - Initial pressure: \(P_1 = 1.960 \mathrm{atm}\) - Final pressure after withdrawing some helium: \(P_2 = 1.710 \mathrm{atm}\) - Withdrawn helium volume at 1.00 atm and 298.15 K: \(V_\text{withdrawn} = 1.75 \mathrm{cm}^{3}\)

Apply the Ideal Gas Law

Since we have two states of the gas (inside and withdrawn), we need to set up two equations applying the Ideal Gas Law: \(P_1V_1 = n_1RT\) and \(P_2V_2 = n_2RT\) Where: - \(P_1\) and \(V_1\) are the initial pressure and the volume of the spherical container - \(P_2\) and \(V_2\) are the final pressure in the container after withdrawing and the remaining volume inside the container - \(n_1\) and \(n_2\) are the initial and final amounts of helium - \(R\) is the ideal gas constant (0.0821 L atm/mol K) - \(T\) is the temperature, which is constant at 298.15 K

Relate withdrawn helium to initial and final states

We are given that the withdrawn helium gas has a volume of \(1.75 \mathrm{cm}^{3}\). Therefore, the difference in volume in the container must also be \(1.75 \mathrm{cm}^{3}\). So we can say that: \(V_1 - V_2 = 1.75 \mathrm{cm}^{3}(1 \frac{\mathrm{mL}}{1\mathrm{cm}^{3}})(\frac{1\mathrm{L}}{1000\mathrm{mL}}) = 0.00175 \mathrm{L}\) This relates the withdrawn helium to the initial and final volume inside the container.

Equate the two Ideal Gas Law equations

Now we can divide the initial Ideal Gas Law equation by the final one to eliminate n1 and n2: \[\frac{P_1V_1}{P_2V_2} = \frac{n_1RT}{n_2RT}\] Simplifying gives: \[\frac{P_1V_1}{P_2V_2} = \frac{n_1}{n_2}\] Now we can substitute our expressions for the volume differences: \[P_1(V_2 + 0.00175) = n_1R\cancel{T}\] and \[P_2V_2 = n_2R\cancel{T}\] Divide these two equations: \[\frac{P_1(V_2 + 0.00175)}{P_2V_2} = \frac{n_1}{n_2}\]

Solve for volume of the spherical container

Now we can substitute the given pressures and solve for \(V_2\): \[\frac{1.960(V_2 + 0.00175)}{1.710V_2} = \frac{n_1}{n_2}\] Solve for \(V_2\): \[V_2(1.960 - 1.710) = 1.960 \times 0.00175\] \[V_2 = \frac{1.960 \times 0.00175}{1.960 - 1.710} = 0.0201 \mathrm{L}\] Now, we can find the initial volume of the spherical container (\(V_1\)): \[V_1 = V_2 + 0.00175 = 0.0201 + 0.00175 = 0.02185 \mathrm{L}\] Thus, the volume of the spherical container is \(0.02185 \mathrm{L}\).

Key concepts

Understanding the Gas Laws

The gas laws are a fundamental aspect of chemistry that provide a quantitative understanding of the behaviors and properties of gases. These laws describe how temperature, volume, and pressure of a gas relate to each other and how they change in response to each other. One of the most vital gas laws is the Ideal Gas Law, which integrates the simple gas laws (Boyle's law, Charles's law, and Avogadro's law) and provides a relationship between pressure (P), volume (V), temperature (T), and number of moles of gas (n). The law is expressed with the equation:

\( PV = nRT \)

Here, R signifies the ideal gas constant. To effectively use this law, as seen in the exercise with the helium gas, we must ensure that the units are consistent, temperatures are in Kelvin, and convert volumes to liters when necessary. By mastering the Ideal Gas Law, we can ascertain unknown properties of a gas if the other properties are provided, as demonstrated in the step-by-step solution to determine the volume of the spherical container.

Chemical Thermodynamics: The Interplay of Energy and Gases

Chemical thermodynamics dives into the energy changes that accompany chemical reactions and physical processes, particularly involving gases in this scenario. It involves concepts like enthalpy, entropy, free energy, and the equilibrium states of chemical processes. Since gases can do work by expanding or compressing, understanding thermodynamics is essential while dealing with gas-based reactions or processes under constant temperature conditions, such as in our helium example.

Thermodynamics tells us that the state of a gas, including its energy, can be altered by its temperature, volume, or the pressure exerted upon it. This knowledge enables us to comprehend how energy is transferred or transformed in a closed system, which is implicit in the use of the Ideal Gas Law, where T maintained constant signifies no change in thermal energy, thus upholding the principles of thermodynamics.

Pressure-Volume Relationship in Gases

The pressure-volume relationship, also known as Boyle's Law, is intrinsic to understanding how gases will react to changes in containment. This law states that for a fixed amount of gas at a constant temperature, the volume of a gas is inversely proportional to its pressure. In other words, if you increase the pressure exerted on a gas, its volume will decrease, provided temperature remains unchanged, and vice versa.

The relationship is described by the equation:

\( P_1V_1 = P_2V_2 \)

This concept is intertwined within the Ideal Gas Law and plays a pivotal role when solving problems related to gas volume and pressure changes, like calculating the final volume of helium remaining in the spherical container after some of it has been withdrawn. An initial understanding of this relationship can significantly simplify tackling complex gas law problems.

Stoichiometry: From Chemical Equations to Gas Quantities

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. In the context of gases, stoichiometry is applied to predict how much gas will be produced or consumed in a reaction, according to the molar ratios outlined in a balanced chemical equation. In processes involving gases where the Ideal Gas Law is used, stoichiometry can determine the amount of gas required or produced at certain conditions of temperature and pressure.

The step-by-step solution of the helium gas problem hints at stoichiometry when it considers the moles of gas before and after some helium is withdrawn. Stoichiometry becomes essential in reactions where gases react in a definite ratio, tying together the mole concept with gas volumes and providing a complete picture of how gases will behave in chemical reactions.