Problem 60 - University Physics with Modern Physics

Question

A flask with a volume of 1.50 \(\mathrm{L}\) , provided with a stopock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at 300 \(\mathrm{K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{Pa}\right) .\) The molar mass of ethane is 30.1 \(\mathrm{g} / \mathrm{mol}\) . The system is warmed to a temperature of \(380 \mathrm{K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

Step-by-Step Solution

Answer

The final pressure is lower than atmospheric, and the mass decreases due to gas escaping during heating.

Step 1: Understand the Initial Setup

Initially, the flask contains ethane gas at a volume of 1.50 L, temperature of 300 K, and pressure of 1.013 \(\times\) 10a{5} Pa. The molar mass of ethane \((\mathrm{C}_{2}\mathrm{H}_{6})\) is 30.1 g/mol. The stopcock is open, meaning the pressure inside is the same as the atmospheric pressure.

Step 2: Use Ideal Gas Law to Find Moles at Initial State

Apply the ideal gas law \(PV = nRT\) to find the initial moles of ethane. Here, \(P = 1.013 \times 10^{5}\,\mathrm{Pa}\), \(V = 1.50\,\mathrm{L} = 1.50 \times 10^{-3}\,\mathrm{m}^3\), \(R = 8.314\,\mathrm{J/(mol\,K)}\), and \(T = 300\,\mathrm{K}\). \[n = \frac{PV}{RT} = \frac{(1.013 \times 10^{5}\,\mathrm{Pa})(1.50 \times 10^{-3}\,\mathrm{m}^3)}{(8.314\,\mathrm{J/(mol\,K)})(300\,\mathrm{K})}\]

Step 3: Simplify and Calculate Moles at Initial State

Simplify the equation and calculate the moles of ethane at the initial state. The calculated moles \(n\) will be used later to determine how much ethane remains in the flask after cooling.

Step 4: Evaluate Moles after Heating and Equilibrium

The flask is heated to 380 K while open to the atmosphere, allowing gas to escape until pressure equilibrates with the external pressure (1.013 \(\times 10^{5}\) Pa). Use the ideal gas law again to find moles at the new temperature. The pressure remains atmospheric as the stopcock was open.

Step 5: Assess Changes after Closing Stopcock and Cooling

After closing the stopcock, cool the flask back to its original temperature of 300 K. The volume of the gas remains constant, but because the number of moles stayed constant after closing the stopcock, use \(PV = nRT\) to find the final pressure considering the cooled temperature.

Step 6: Calculate Final Pressure

Rearrange the ideal gas law to find final pressure: \[ P_{\text{final}} = \frac{nRT_{\text{final}}}{V} \] substitute the values with the final temperature \(T_{\text{final}} = 300\,\mathrm{K}\) and the volume 1.50 \( \times 10^{-3} \mathrm{m}^3 \).

Step 7: Determine Mass of Ethane Remaining

With the final moles \(n\) calculated from after the heating process, compute the mass by multiplying moles by the molar mass of ethane. Mass \(= n \times 30.1\,\mathrm{g/mol}\).

Key Concepts

Moles of Gas

The concept of moles is essential when dealing with gases in chemical reactions and calculations. A mole is a unit that measures the amount of substance, and one mole encompasses Avogadro's number, approximately \(6.022 \times 10^{23}\) entities, like atoms or molecules.

When applying the ideal gas law, represented by \(PV = nRT\), the moles of gas \(n\) can be determined from the known values of pressure \(P\), volume \(V\), and temperature \(T\) with the gas constant \(R\) considered.

• \(P\) stands for pressure.
• \(V\) denotes volume.
• \(T\) is temperature.
• \(R\) is the ideal gas constant, \(8.314\,\mathrm{J/(mol\,K)}\).

By rearranging the ideal gas law to solve for \(n\), you get \(n = \frac{PV}{RT}\). This formula allows the calculation of the amount of gas, in moles, present under specific conditions.

Pressure Equilibrium

Understanding pressure equilibrium is crucial when dealing with gases under changing conditions. Pressure equilibrium occurs when the pressure of gas inside a container matches the external pressure. In practical terms, if a container, such as a flask, is open to the atmosphere and then closed when heated, it will reach an equilibrium with the outside air pressure.

In our exercise, when the ethane gas is warmed to 380 K with an open stopcock, the pressure inside equilibrates with the atmospheric pressure. This means no net gas movement occurs across the boundary once equilibrium is reached.

When equilibrium is achieved, the moles of a gas in the flask can be reassessed at the new temperature while maintaining a stable pressure. This ensures that when the stopcock is closed, that same pressure equilibrium defines the system's state even after subsequent cooling.

Molar Mass Calculation

Molar mass is a property that indicates the mass of one mole of a substance, typically measured in grams per mole (g/mol). It plays a vital role in converting between the mass of a substance and its corresponding amount in moles.

For ethane, \((C_2H_6)\), the molar mass is provided as 30.1 g/mol. This means that one mole of ethane weighs 30.1 grams.

To determine how much ethane remains in the flask after undergoing the temperature changes, you can calculate the mass by simply multiplying the moles of ethane that remain by the molar mass using the formula:

• Mass = moles \(n\) \(\times\) molar mass \(30.1\,\mathrm{g/mol}\)

This calculation plays a crucial part in the exercise by determining the final amount of ethane left under the given conditions.

Temperature Change in Gases

Temperature changes impact gases significantly by affecting their pressure, volume, or both. The ideal gas law offers a framework to understand these effects. In this situation, when the temperature of the ethane gas increases from 300 K to 380 K, the gas expands due to increased kinetic energy of the gas molecules, if the volume is allowed to change.

However, with the volume fixed in the exercise, and the stopcock opened to allow equilibrium, the system adjusts by maintaining constant pressure while temperature changes. Consequently, a portion of the gas escapes, reducing the number of moles.

Upon cooling the gas back to 300 K with the stopcock closed, you use the reduced number of moles from the equilibrium state to re-calculate the pressure and final states reflecting the influence of the initial temperature change on the gas properties. This demonstrates the gas's sensitivity to temperature, crucial to predicting changes within closed or semi-closed systems.