Problem 127
Question
A container of gas has a pressure of \(550 .\) Torr. A chemical change then occurs that consumes half of the molecules present at the start and produces two new molecules for each three consumed. Calculate the new pressure in the container if \(T\) and \(V\) are unchanged.
Step-by-Step Solution
Verified Answer
The new pressure is approximately 458.33 Torr.
1Step 1: Understand the Initial Conditions
We are given an initial pressure of the gas in the container, which is \(550\) Torr. We know temperature \((T)\) and volume \((V)\) remain constant throughout the process.
2Step 2: Analyze the Chemical Change
The chemical change consumes half of the molecules initially present, and in their place, two new molecules are produced for every three consumed. Essentially, for every one molecule consumed, \(\frac{2}{3}\) molecules are produced.
3Step 3: Calculate the Change in Molecules
Initially, assume there are \(n\) molecules. Half of these, \(\frac{n}{2}\), are consumed. According to the problem, for each three consumed, two new are produced, so \(\frac{1}{2} n \times \frac{2}{3} = \frac{1}{3} n\) new molecules are produced.
4Step 4: Determine the New Total Number of Molecules
The original number of molecules is \(n\). After the chemical change, the number of molecules is given by the original minus those consumed plus those produced: \(n - \frac{1}{2}n + \frac{1}{3}n = \frac{5}{6}n\).
5Step 5: Apply the Ideal Gas Law
Using the relationship from the ideal gas law \(PV = nRT\), and noting that \(T\) and \(V\) are constant, pressure \(P\) is proportional to the number of moles \(n\). The new pressure \(P_{\text{new}}\) is \( \frac{5}{6}\) of the initial pressure \(P_{\text{initial}} = 550\text{ Torr}\).
6Step 6: Calculate the New Pressure
Compute the new pressure: \(P_{\text{new}} = \frac{5}{6} \times 550 = 458.33\text{ Torr}\) after rounding to two decimal places.
Key Concepts
Understanding Chemical Change in Gas ReactionsInsights into the Ideal Gas Law and Its ApplicationPressure Calculation after Chemical Reaction
Understanding Chemical Change in Gas Reactions
Chemical change involves a transformation that alters the composition of molecules in a substance. In gas reactions, chemical changes impact the number and type of gas molecules present in a container. They are crucial because they help us understand reactions and their impacts, such as changes in pressure and volume.
Let's consider the exercise, where initially we have a specific number of gas molecules. During the chemical change, half of these molecules are consumed, and to replace every three consumed molecules, two new ones are created. This effectively alters the quantity of molecules.
Understanding this transformation helps dictate how to calculate the resulting pressure, utilizing concepts like stoichiometry, which explains the ratios of reactants and products. By understanding these ratios, we predict the future state of the gas mixture.
Let's consider the exercise, where initially we have a specific number of gas molecules. During the chemical change, half of these molecules are consumed, and to replace every three consumed molecules, two new ones are created. This effectively alters the quantity of molecules.
Understanding this transformation helps dictate how to calculate the resulting pressure, utilizing concepts like stoichiometry, which explains the ratios of reactants and products. By understanding these ratios, we predict the future state of the gas mixture.
- Original molecules: Start with number, say, \( n \).
- Half consumed: \( \frac{1}{2}n \).
- New molecules produced: \( \frac{2}{3}\) molecules for every 3 consumed.
Insights into the Ideal Gas Law and Its Application
The Ideal Gas Law is a fundamental principle of chemistry expressed by the formula \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is temperature. This law allows us to predict how a gas behaves under various conditions.
In the given problem, since temperature \( T \) and volume \( V \) remain constant, pressures can be directly compared using moles \( n \). This presents pressure as directly proportional to the number of molecules.
When there's a chemical change like in our exercise, where the number of molecules changes, you use the initial pressure and adjust it based on the fraction of molecules remaining and produced.
In the given problem, since temperature \( T \) and volume \( V \) remain constant, pressures can be directly compared using moles \( n \). This presents pressure as directly proportional to the number of molecules.
When there's a chemical change like in our exercise, where the number of molecules changes, you use the initial pressure and adjust it based on the fraction of molecules remaining and produced.
- Initial pressure is given.
- Apply the change in moles to adjust pressure.
Pressure Calculation after Chemical Reaction
To calculate pressure changes during a chemical reaction, an understanding of how gas molecules interact according to the ideal gas law is necessary. In our exercise, given initial conditions of 550 Torr pressure, we determine the impact of a chemical change that affects molecule numbers.
This calculation starts with adjusting the number of moles, as shown in the step-by-step solution:
The equation reveals a new pressure of \( 458.33 \) Torr, ensuring consistent application of the ideal gas law principles.
This reliable calculation proves useful not just for prediction, but for practical applications in various scientific and industrial settings.
This calculation starts with adjusting the number of moles, as shown in the step-by-step solution:
- The total number of molecules after reaction: \( \frac{5}{6}n \).
- Original pressure \( 550 \) Torr; new pressure relates to this proportionally.
The equation reveals a new pressure of \( 458.33 \) Torr, ensuring consistent application of the ideal gas law principles.
This reliable calculation proves useful not just for prediction, but for practical applications in various scientific and industrial settings.
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