Problem 62 - University Physics with Modern Physics

Question

A flask with a volume of \(1.50 \mathrm{L},\) provided with a stop cock, 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 \(490 \mathrm{K},\) with the stop cock 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

(a) The final pressure is approximately 0.94 x 10^5 Pa. (b) 1.84 g of ethane remains in the flask.

Step 1: Understanding the Problem

We are given initial conditions for a flask containing ethane gas and asked to determine the final pressure and mass of ethane after adjusting the temperature. The process involves warming with the stopcock open (gas exits until pressure equalizes with external pressure) and then cooling with the stopcock closed.

Step 2: Ideal Gas Law Calculation

Initially, the gas is subjected to the atmospheric pressure and occupies a volume of 1.50 L at 300 K. Using the ideal gas law formula \( PV = nRT \), calculate the initial number of moles (\( n \)) in the flask: \( n = \frac{PV}{RT} \). Given \( P = 1.013 \times 10^5 \text{ Pa} \), \( V = 1.5 \times 10^{-3} \text{ m}^3 \), and \( R = 8.314 \text{ J/mol K} \), we find \( n \approx 0.061 \) mol.

Step 3: Gas Escapes During Heating

When the system is heated to 490 K with the stopcock open, the gas inside expands and escapes until the pressure inside the flask equals the atmospheric pressure (1.013 x 10^5 Pa). According to Boyle's Law (for constant temperature processes), since the volume is constant, the moles adjust to maintain pressure.

Step 4: Number of Moles after Heating

At 490 K (with stopcock open), pressure equals atmospheric pressure; therefore, when cooled with the stopcock closed, the number of moles that remains in the flask is based on the temperature and initial conditions. Use the ideal gas law with the new volume (same as before) and original pressure to determine the new moles \( n = \frac{PV}{RT} = \frac{1.013 \times 10^5 \times 1.5 \times 10^{-3}}{8.314 \times 300} \approx 0.061 \) mol.

Step 5: Final Pressure Calculation

With known moles remaining at the original temperature of 300 K, apply the ideal gas law to find the final pressure: \( P = \frac{nRT}{V} = \frac{0.061 \times 8.314 \times 300}{1.5 \times 10^{-3}} \approx 0.94 \times 10^5 \text{ Pa}.\)

Step 6: Mass of Ethane Remaining

To find the mass of ethane remaining, multiply the moles by the molar mass: \( ext{mass} = n \times ext{molar mass} = 0.061 \times 30.1 = 1.84 \text{ g}. \)

Key Concepts

Boyle's Law and Its Role in Gas Behavior

Boyle's Law is a key concept in understanding how gases behave under constant temperature conditions. It states that the pressure of a gas is inversely proportional to its volume, as long as the temperature remains constant. This means if you increase the volume, the pressure decreases, and vice versa.
In the exercise given, when the flask is warmed to 490 K with the stopcock open, the gas expands. According to Boyle's Law, the volume remains constant since it's a flask, but the gas escapes to equalize with atmospheric pressure, which is 1.013 x 10⁵ Pa. Boyle's Law helps us understand this process because the changes in pressure and gas escaping are directly related to the volumes available inside and outside the flask.
Once the stopcock is closed and the flask begins to cool back to its original temperature, Boyle’s Law is not directly applied anymore since the conditions then involve changing temperatures. However, the initial escape of gas under constant temperature conditions can be best explained with Boyle’s Law, as it lays the groundwork for understanding how pressure-volume relationships adjust.

Understanding Gas Pressure Calculation

Gas pressure calculations are central to analyzing the behavior of gases in enclosed systems. The Ideal Gas Law, expressed as \( PV = nRT \), is essential for calculating the relationships between pressure (P), volume (V), temperature (T), and the amount (n) or moles of gas, where \( R \) is the ideal gas constant.
In the given exercise, we initially use this formula to determine the number of moles of ethane present at the original conditions: \( n = \frac{PV}{RT} \). Using \( P = 1.013 \times 10^5 \) Pa, \( V = 1.5 \times 10^{-3} \) m³, \( R = 8.314 \) J/mol K, and \( T = 300 \) K, we find that the moles of ethane initially present equal approximately 0.061 mol.
After the warming and gas escape with the stopcock open, the pressure is re-calculated once the system returns to its original temperature of 300 K but with decreased moles remaining. The same ideal gas law returns a final pressure of about 0.94 x 10⁵ Pa. This shows how gas pressure changes based on moles present and the constant volume of the system.

Calculating Molar Mass and Its Significance

The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole. For ethane \((\text{C}_2\text{H}_6)\), the given molar mass is 30.1 g/mol, which is useful for converting between the amount of substance in moles and its mass.
To find the mass of ethane remaining after the sequence of heating with the stopcock open, then cooling while closed, we use the remaining moles determined from the gas law calculations. The final step in the exercise involves multiplying the moles by the molar mass: \( \text{mass} = n \times \text{molar mass} = 0.061 \times 30.1 = 1.84 \) grams.
This computation shows how molar mass connects the amount of substance in chemical terms (moles) to a practical measurement (grams) that can be weighed. Understanding molar mass is crucial for translating theoretical chemistry calculations into laboratory and real-world applications.