Problem 8 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Two bulbs \(A\) and \(B\), with volumes \(V_{A}=1.00 \mathrm{dm}^{3}\) and \(V_{B}\) \(=5.00 \mathrm{dm}^{3},\) are connected via a tap. The volume of the connecting tubing is negligible. Bulb A contains gas at a pressure of 6.00 bar while bulb \(\mathrm{B}\) contains a vacuum. (a) The temperature of the whole apparatus is maintained at \(298 \mathrm{K}\). If the tap is opened, calculate the pressure of gas in the system after opening the tap. (b) The tap is closed and bulb \(\mathrm{B}\) is then immersed in an oil bath at a temperature of \(423 \mathrm{K}\) while the temperature of bulb \(\mathrm{A}\) is maintained at \(298 \mathrm{K}\). Calculate the resulting pressures in each bulb. (c) The tap is opened again. What is the final pressure and the number of moles of gas in each bulb? (Section 8.2)

Step-by-Step Solution

Answer

(a) 1.00 bar (b) A: 1.00 bar, B: 1.72 bar (c) 1.20 bar, 0.242 mol each.

Step 1: Calculate Initial Moles in Bulb A

Using the ideal gas equation, which states \(PV = nRT\) (where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature), we first calculate the number of moles in bulb A. Given:\[ P = 6.00 \text{ bar} = 6.00 \times 10^5 \text{ Pa}, \quad V = 1.00 \text{ dm}^3 = 1.00 \times 10^{-3} \text{ m}^3, \quad T = 298 \text{ K}, \quad R = 8.31 \text{ J/mol}\cdot\text{K} \] We solve for \(n\):\[ n = \frac{PV}{RT} = \frac{6.00 \times 10^5 \times 1.00 \times 10^{-3}}{8.31 \times 298} = 0.242 \text{ mol} \]

Step 2: Calculate Pressure After Opening the Tap (Part a)

When the tap is opened, gas from bulb A spreads into bulb B. The total volume becomes \(V_A + V_B = 1.00 + 5.00 = 6.00 \text{ dm}^3\). The pressure of the system is then calculated using: \[ P_{final} = \frac{nRT}{V_{total}} = \frac{0.242 \times 8.31 \times 298}{6.00 \times 10^{-3}} = 1.00 \text{ bar} \] So the pressure after opening the tap is 1.00 bar.

Step 3: Calculate Pressures with Bulb B at High Temperature (Part b)

After closing the tap, bulb B is heated to \(T_B = 423 \text{ K}\) while bulb A remains at 298 K. As volumes don't change, initial and final states for each bulb separately are considered. For bulb A, the number of moles stays the same (0.242 mol) at 298 K, keeping pressure at 1.00 bar. For bulb B, using the ideal gas law: \[ P_BV_B = n_BRT_B \Rightarrow P_B = \frac{n_BRT_B}{V_B} = \frac{0.242 \times 8.31 \times 423}{5.00 \times 10^{-3}} \approx 1.72 \text{ bar} \] The pressure in bulb B becomes approximately 1.72 bar.

Step 4: Final Pressure and Moles on Re-Opening the Tap (Part c)

When the tap reopens, gases exchange so that the system equilibrates again to a uniform pressure. Using combined moles \([0.242 (A) + 0] \text{ mol} = 0.242 \text{ mol}\) and \(T_{avg} = \frac{298 + 423}{2} = 360.5 \text{ K}\), we find: \[ P_{final} = \frac{nRT_{avg}}{V_{total}} = \frac{0.242 \times 8.31 \times 360.5}{6.00 \times 10^{-3}} \approx 1.20 \text{ bar} \] Number of moles remains unchanged at \(0.242 \text{ mol}\). Both bulbs will have the same final pressure.

Key Concepts

Gas Pressure Calculations

Understanding how to calculate gas pressure is vital when dealing with gases in closed containers. The pressure of a gas is determined by how many molecules collide with the walls of its container. An increase in the number of these collisions raises the pressure. In thermodynamics, when using the Ideal Gas Law, you express this relationship mathematically: \[ P = \frac{nRT}{V} \] Here, \( P \) is pressure, \( n \) is the number of moles, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the container. When solving pressure problems:

• Ensure units are consistent you typically convert pressure to Pascal and volume to cubic meters.
• Use a constant temperature scale, preferably Kelvin, to maintain linearity in calculations.
• If the volume increases or decreases without changing temperature, the pressure will inversely change according to Boyle's Law.

By applying these steps systematically, you can find the unknown pressure in varied scenarios like the one in the exercise.

Thermodynamics

Thermodynamics is the branch of physics concerned with heat and temperature and their relation to energy and work. Central to this exercise is how temperature changes affect gas behavior in confined systems. When you introduce a temperature change:

• The energy of gas particles increases with a temperature rise, prompting faster movement and increased pressure if the volume remains constant.
• Conversely, cooling gas reduces molecular energy movement, decreasing pressure if the volume stays the same.
• For isolated systems, such as bulb A and B, you observe changes using laws like Charles' Law and Gay-Lussac's Law, connecting temperature and pressure or temperature and volume under constant conditions.

In practical terms:

• Raising the temperature in bulb B from 298 K to 423 K increases the pressure due to faster moving particles.
• Spreading out these molecules into a larger volume by opening the tap redistributes this pressure, finding equilibrium between both containers.

Understanding these principles helps you predict and control pressure changes reliably when manipulating gas systems in thermodynamics.

Volume and Temperature Relation

The relationship between volume and temperature in gases is a fundamental aspect of thermodynamics, often explored through Charles' Law. Charles' Law states that, at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. This means: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Here's what you need to know:

• A temperature increase can cause gas expansion if the container allows for volume change, which is often visualized in tasks like inflating a balloon.
• Conversely, if a gas is heated in a rigid, unchangeable volume, pressure increases as molecules push more forcefully against their container.

In vacuum environments, such as bulb B before being heated, gas occupies any available space. When the tap opens back to bulb A, there's a rebalancing act, ensuring all molecules occupy the newly available collective volume (bulb A + bulb B) as they share this space. It's crucial to maintain temperature in one isolated part of the system to judge effects more clearly, showing the elegant interplay between temperature, volume, and pressure.