Question

Pressure of \(1 \mathrm{~g}\) of an ideal gas \(\mathrm{A}\) at \(27^{\circ} \mathrm{C}\) is found to be 2 bar. When \(2 \mathrm{~g}\) of another ideal gas \(\mathrm{B}\) is introduced in the same flask at same temperature the pressure becomes 3 bar. What is the relationship between their molecular masses? (a) \(2 M_{\mathrm{B}}=M_{\mathrm{A}}\) (b) \(M_{\mathrm{B}}=M_{\mathrm{A}}\) (c) \(M_{\mathrm{B}}=4 M_{\mathrm{A}}\) (d) \(M_{\mathrm{B}}=2 M_{\mathrm{A}}\)

Short answer

The relationship between the molecular masses of the two gases is given by: (b) \(M_{\mathrm{B}} = M_{\mathrm{A}}\).

Step-by-step solution

Assumptions using Ideal Gas Law

Assume the volume V of the flask and the ideal gas constant R to be the same for both gases. According to the Ideal Gas Law, the pressure P of a gas is directly proportional to the number of moles n and the temperature T, and inversely proportional to the volume V. Since the temperature and volume are constant for both gases, the pressure is directly proportional to the number of moles: P = (n*R*T)/V.

Calculate moles of gas A

Using the relation from the ideal gas law and given that the pressure of 1 g of gas A is 2 bar, calculate the amount in moles of gas A using its molecular mass MA and the Avogadro's number. The number of moles nA = mass of A / molecular mass of A = (1 g) / MA.

Calculate total moles after adding gas B

When gas B is introduced, the total pressure becomes 3 bar which is the sum of the partial pressures of gases A and B. Since pressure is directly proportional to the number of moles, the partial pressure of B will be 1 bar (3 bar - 2 bar for A). Calculate the number of moles of B, nB = mass of B / molecular mass of B = (2 g) / MB.

Use Dalton's Law of Partial Pressures to relate pressures

According to Dalton's Law, the total pressure is the sum of the partial pressures of the individual gases. Therefore, PA + PB = Total P, where PA = 2 bar (1 mole of A's partial pressure) and PB = the resulting pressure after adding B. Relate the number of moles of A and B with their respective pressures assuming the volume and temperature remain constant.

Derive the relationship between MA and MB

The partial pressure of gas A (PA = 2 bar) is due to 1 g of A, and the partial pressure of gas B (PB = 1 bar) is due to 2 g of B. Using the proportionality from the ideal gas law, the ratio of their pressures is equal to the ratio of moles, so PA/PB = nA/nB. This ratio is (2 bar / 1 bar) = ((1 g / MA) / (2 g / MB)). Simplify and solve for the relationship between MA and MB.

Evaluate the Molecular Mass Relationship

Solving the equation from the previous step yields the relationship 2/1 = (1 g / MA) / (2 g / MB) or 2 * MA = MB. Rearranging gives MA = 2 * MB or MB = 2 * MA. Therefore, the molecular mass of B is twice that of A.

Key concepts

Dalton's Law of Partial Pressures

In an environment with a mixture of gases, each gas contributes to the total pressure as if it were the only gas present. This is the essence of Dalton's Law of Partial Pressures, a principle that is critical to solving problems involving gas mixtures.

Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Mathematically, it is expressed as:
\[ P_{\text{total}} = P_1 + P_2 + \cdots + P_n \]
where \( P_{\text{total}} \) is the total pressure, and \( P_1, P_2, \ldots, P_n \) are the partial pressures of the gases.

This law is fundamental when determining the amount of each gas in a mixture. To illustrate, in the given exercise, we first find the pressure of gas A and then, upon introducing gas B, we apply Dalton's Law to understand the effects on total pressure. Here, the presence of two different gases leads us to determine the partial pressure each gas contributes, which in turn helps us to deduce the molecular mass relationship between the gases. Dalton's Law allows us to isolate the impact of each gas and better analyze the properties of the gas mixture.

Molecular Mass Relationship

Understanding the molecular mass relationship between different gases is another central concept in chemistry, essential for both practical laboratory work and theoretical studies.

The molecular mass, essentially the mass of one molecule of a substance, is often expressed in atomic mass units (amu). The relationship between the masses of different molecules impacts how we calculate the number of moles, which in turn affects the ideal gas law calculations.

In our exercise, the relationship between the molecular mass of gases A (\( M_A \)) and B (\( M_B \)) is uncovered by examining the ratio of their pressures and masses. It demonstrates how the molecular mass can be inferred by combining the information provided by both Dalton's Law of Partial Pressures and the Ideal Gas Law. Knowing one gas's mass allows us to calculate the other's if we understand their relationship, illustrating the interconnected nature of these formulas and the importance of molecular mass in gas behavior analysis.

Avogadro's Number

Avogadro's number, denoted as \( N_A \), is a fundamental constant in chemistry that represents the number of constituent particles (usually atoms or molecules) present in one mole of a substance. Its approximate value is \( 6.022 \times 10^{23} \) particles per mole.

This immense number allows chemists to correlate the microscopic world of atoms and molecules with the macroscopic world that we can measure by linking the amount of substance to its mass. For instance, in the context of the ideal gas law problems, Avogadro's number enables us to convert between mass and moles, which is essential for calculating the pressure of a gas, as seen in our exercise.

When calculating the number of moles of gas A or B, we implicitly rely on Avogadro's constant to establish a direct connection between the microscopic property (molecular mass) and the macroscopic measurement (pressure). Through Avogadro's number, subtle aspects of the atomic-scale behavior influence the tangible and observable properties of gases.