Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

A gas cylinder has \(370 \mathrm{~g}\) of oxygen at \(298 \mathrm{~K}\) and 30 atm pressure. If the cylinder was heated upto \(348 \mathrm{~K}\) then the valve were held open until the gas pressure was 1 atm and the temperature remains \(348 \mathrm{~K}\). What mass of oxygen would escape in this condition? (a) \(349 \mathrm{~g}\) (b) \(359 \mathrm{~g}\) (c) \(329 \mathrm{~g}\) (d) \(339 \mathrm{~g}\)

Step-by-Step Solution

Answer

The mass of oxygen that escaped is 359 g.

Step 1: Identify the given values

The problem provides us with the following values:- Initial mass of oxygen: 370 g- Initial temperature, \(T_1 = 298 \text{ K}\)- Initial pressure, \(P_1 = 30 \text{ atm}\)- Final temperature, \(T_2 = 348 \text{ K}\)- Final pressure, \(P_2 = 1 \text{ atm}\)We are asked to find out how much mass of oxygen escaped after changing conditions to \(T_2\) and \(P_2\).

Step 2: Determine initial moles using Ideal Gas Law

First, let's determine the initial moles of gas using the ideal gas law, \(PV = nRT\).For initial conditions:\[P_1V = n_1RT_1\]\(n_1\) is the initial moles of oxygen. The molecular weight of \(O_2\) is 32 g/mol, so the initial moles are:\[n_1 = \frac{370 \text{ g}}{32 \text{ g/mol}} \approx 11.56 \text{ mol}\]

Step 3: Apply Ideal Gas Law at final conditions

For final conditions, the relationship is:\[P_2V_f = n_2RT_2\]Since \(P_2 = 1 \text{ atm}\) and \(T_2 = 348 \text{ K}\), you can write:\[n_2 = \frac{P_2V}{RT_2}\]Replacing the variables with those at initial and final conditions, note that the volume \(V\) does not change:\[\frac{n_1RT_1}{P_1} = \frac{n_2RT_2}{P_2}\]

Step 4: Calculate final moles after conditions change

Reorganize the equation and solve for final moles \(n_2\):\[n_2 = n_1 \times \frac{P_2}{P_1} \times \frac{T_1}{T_2}\]Plug in the numbers:\[n_2 = 11.56 \text{ mol} \times \frac{1}{30} \times \frac{298}{348} \approx 0.329 \text{ mol}\]

Step 5: Calculate the mass of gas that escaped

Find the difference in moles to see how many moles escaped:\[\text{Moles of gas escaped} = n_1 - n_2 = 11.56 - 0.329 = 11.231 \text{ mol}\]Convert this to grams:\[\text{Mass of gas escaped} = 11.231 \text{ mol} \times 32 \text{ g/mol} \approx 359.392 \text{ g}\]Round to the nearest gram to match the answer choices: 359 g.

Key Concepts

Moles of gas

When studying gases, the concept of moles plays an essential role. A mole is a unit used to calculate the amount of substance. It helps to relate the mass of a substance to the number of particles, such as atoms or molecules, in measured amounts.
This is particularly useful in gas calculations where volume, temperature, and pressure are involved, and there is a need to interconvert between mass, moles, and volume using the Ideal Gas Law.
For oxygen gas, the molecular formula is \(O_2\), meaning each molecule consists of two oxygen atoms. The molecular weight of \(O_2\) is 32 grams per mole. In calculations like these, knowing the molecular weight is necessary to convert grams of gas to moles and vice versa.

• The formula to find moles \(n\) from mass \(m\) is: \(n = \frac{m}{M_{molar}}\), where \(M_{molar}\) is the molar mass.
• This step is crucial as it transforms a real-world amount into a scientific measurement convenient for problem solving.

Understanding moles lays the groundwork for applying the Ideal Gas Law to solve problems regarding changes in conditions of a gas system.

Temperature and pressure changes

Temperature and pressure are fundamental properties that affect a gas's behavior. When addressing gases, we often refer to the Ideal Gas Law, represented as \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for moles, \(R\) for the gas constant, and \(T\) for temperature. This law is crucial in connecting the physical quantities of a gas.
When any two of the properties among pressure, temperature, and volume change while the moles stay constant, the Ideal Gas Law can predict and determine the new conditions.

• The relationship is consistent if the temperature is in Kelvin and the pressure is in the appropriate units.
• In problems where temperature increases, you can expect pressure or volume to change if moles remain constant.
  

Adjusting the equation to reflect conditions before and after a change provides insight into how gases behave under varying conditions. This is illustrated by using ratios, such as \(\frac{T_1}{T_2}\) and \(\frac{P_2}{P_1}\), reflecting changes from initial to final states.

Oxygen gas calculations

Oxygen gas calculations often require a blend of knowledge about molecular properties and laws governing gases. In the given exercise, we are dealing with a dynamic situation where temperature and pressure changes lead to some amount of oxygen escaping.

• To find out how much oxygen escapes, we use initial and final conditions plugged into the Ideal Gas Law.
• The difference in moles from initial (before heating or opening the valve) to final (after reaching the new temperature and pressure) gives the quantity of substance that escapes.
  
• Conversion back to mass gives practical, understandable results. Knowing the molecular weight of \(O_2\) (32 g/mol) allows us to easily switch from the abstract (moles) to the concrete (grams).

In this problem, understanding these concepts helps us determine that approximately 359 grams of oxygen escape under the new conditions. It demonstrates the practical application of chemistry concepts in real-world scenarios.