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A flask with a volume of \(1.50 \mathrm{~L}\), provided with a stopcock, 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?

Short Answer

Expert verified
(a) The final pressure is \(8.047 \times 10^4 \text{ Pa}\). (b) The remaining ethane mass is 1.45 g.

Step by step solution

01

Understand the Problem Setup

At the beginning, we have a flask with ethane gas at a certain temperature and atmospheric pressure. The flask is heated, resulting in some gas escaping. Later, it's cooled back to the original temperature with the stopcock closed. We need to find the final pressure inside the flask and the mass of ethane that remains.
02

Determine the Initial Moles of Ethane

Use the ideal gas law to calculate the initial moles of ethane (\(n_i\)) in the flask. The ideal gas law is \( PV = nRT \), where \(P = 1.013 \times 10^5 \text{ Pa}\), \(V = 1.50 \text{ L} = 1.50 \times 10^{-3} \text{ m}^3\), \(R = 8.314 \text{ J/mol K}\), and \(T = 300 \text{ K}\).Solving for \(n_i\):\[ n_i = \frac{PV}{RT} = \frac{1.013 \times 10^5 \times 1.50 \times 10^{-3}}{8.314 \times 300} \approx 0.0609 \text{ mol} \]
03

Apply Gas Law During Heating

When the flask is heated to 380 K with the stopcock open, some ethane escapes to maintain atmospheric pressure. The pressure inside remains \(1.013 \times 10^5 \text{ Pa}\) while heated. Since volume is constant, we use \( \frac{n_f}{n_i} = \frac{T_i}{T_f} \) to find moles remaining \(n_f\):\[ n_f = n_i \times \frac{T_i}{T_f} = 0.0609 \text{ mol} \times \frac{300}{380} \approx 0.0481 \text{ mol} \]
04

Calculate Pressure on Cooling

After cooling back to 300 K with the stopcock closed, we apply the ideal gas law to find the final pressure \(P_f\).\[ P_f = \frac{n_fRT}{V} = \frac{0.0481 \times 8.314 \times 300}{1.50 \times 10^{-3}} \approx 8.047 \times 10^4 \text{ Pa} \]
05

Calculate Mass of Remaining Ethane

Determine the mass of ethane using the remaining moles. With a molar mass of 30.1 g/mol:\[ m = n_f \times M = 0.0481 \text{ mol} \times 30.1 \text{ g/mol} \approx 1.45 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ethane Gas
Ethane gas, represented by the chemical formula \(\text{C}_2\text{H}_6\), is a simple hydrocarbon belonging to the alkane family. This compound consists of two carbon atoms bonded together and surrounded by six hydrogen atoms. Ethane is a gas at standard temperature and pressure (STP) and is a major component of natural gas. It is colorless and odorless, and it undergoes combustion to be used as a fuel source. In chemistry and physics, understanding gas behavior involves studying its properties, like pressure, volume, and temperature. This forms the basis of our problem, where ethane undergoes changes in these conditions as it is heated and cooled in a closed container. The Ideal Gas Law is crucial in analyzing such scenarios, providing insights into gas behavior in confined scenarios.
Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance. For ethane, the molar mass is given as \(30.1 \text{ g/mol}\). It is calculated by adding the atomic masses of all the atoms in one molecule of the compound.
  • Each carbon atom has an atomic mass of approximately \(12.01\text{ g/mol}\).
  • Each hydrogen atom has an atomic mass of approximately \(1.008\text{ g/mol}\).
Therefore, the molar mass of ethane is found by adding up the atomic masses of 2 carbon atoms and 6 hydrogen atoms: \[2 \times 12.01 \text{ g/mol} + 6 \times 1.008 \text{ g/mol} = 30.1 \text{ g/mol}.\]This value is used to convert between the amount of substance in moles and its mass in grams, as shown in the solution where the mass of ethane is calculated from the number of moles present after heating and cooling.
Pressure Calculation
Pressure is the measure of force exerted per unit area, and in the context of gases, it often refers to the pressure exerted by gas particles colliding against the walls of their container. In this exercise, ethane gas is initially at atmospheric pressure, given as \(1.013 \times 10^5\text{ Pa} (\text{Pascals})\).The Ideal Gas Law (\[ PV = nRT \] ) is pivotal to calculate the pressures when the temperature and moles change. This equation relates pressure (\(P\)), volume (\(V\)), the number of moles (\(n\)), the gas constant (\(R = 8.314\text{ J/mol K}\)), and temperature in Kelvin (\(T\)).
  • Initially, pressure matches atmospheric conditions as the gas is at \(300\text{ K}\).
  • When temperature rises, some gas escapes until pressure balances again with atmospheric levels.
  • Upon cooling with a closed stopcock, the decreased moles lead to a lower pressure, calculated using the ideal gas law again.
Temperature Change
Temperature represents the average kinetic energy of particles within a substance. In the ideal gas model, when the temperature increases, the kinetic energy of the gas particles also increases, causing them to move more rapidly. Conversely, a decrease in temperature reduces the energy and speed of the particles.In this exercise, temperature changes play a crucial role:
  • The initial temperature of ethane is \(300\text{ K}\), which sets the starting point for the calculations.
  • On heating to \(380\text{ K}\), ethane expands, and some gas particles escape to maintain equilibrium at atmospheric pressure once the stopcock is opened.
  • After closing the system, cooling it back to the initial temperature affects the remaining gas quantity, which is vital for determining the final pressure inside the flask.
Understanding temperature change helps in predicting the behavior of gas under varying thermal conditions, which is essential for accurate pressure and volume calculations.

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