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Chapter 5: Problem 22

A gas occupies \(56.44 \mathrm{L}\) at 2.00 atm and \(310 .\) K. If the gas is compressed to \(23.52 \mathrm{L}\) and the temperature is lowered to \(281 \mathrm{K},\) what is the new pressure?

Short Answer

Expert verified
The new pressure is approximately 4.35 atm.

Step by step solution

01

Understand the Ideal Gas Law

The Ideal Gas Law is given by \( PV = nRT \), but in this problem, we do not focus on the amount of gas \( n \) or the gas constant \( R \). Instead, we need to use the combined gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \), which allows us to solve for a change in conditions of a gas.
02

Identify Initial Conditions

Identify the initial conditions from the problem: Pressure \( P_1 = 2.00 \) atm, Volume \( V_1 = 56.44 \) L, and Temperature \( T_1 = 310 \) K.
03

Identify Final Conditions

Identify the final conditions from the problem: Volume \( V_2 = 23.52 \) L and Temperature \( T_2 = 281 \) K. We need to find the final Pressure \( P_2 \).
04

Rearrange the Combined Gas Law

Rearrange the combined gas law equation to solve for \( P_2 \): \( P_2 = P_1 \cdot \frac{V_1}{V_2} \cdot \frac{T_2}{T_1} \).
05

Substitute Values and Solve

Substitute the known values into the rearranged equation: \( P_2 = 2.00 \cdot \frac{56.44}{23.52} \cdot \frac{281}{310} \). Calculate to find \( P_2 \).
06

Calculate the Final Pressure

Perform the calculations: \( \frac{56.44}{23.52} \approx 2.40 \) and \( \frac{281}{310} \approx 0.906 \). Therefore, \( P_2 = 2.00 \cdot 2.40 \cdot 0.906 \approx 4.35 \) atm.

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Key Concepts

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

Combined Gas Law
The Combined Gas Law is a powerful tool in understanding how gases behave when two or more conditions change simultaneously. It incorporates Boyle's Law, Charles's Law, and Gay-Lussac's Law. The formula for the Combined Gas Law is: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]This equation makes it possible to predict how changes in pressure, volume, and temperature affect a gas, assuming a fixed amount of gas is present. Here鈥檚 how each part contributes:
  • Boyle's Law: Shows how pressure and volume are inversely proportional when temperature is constant.
  • Charles's Law: Demonstrates the direct proportionality between volume and temperature at constant pressure.
  • Gay-Lussac's Law: Highlights the direct proportionality between pressure and temperature when volume is constant.
In practical use, the Combined Gas Law helps predict the state of a gas after undergoing several changes at once.
Pressure Calculation
To calculate pressure changes using the Combined Gas Law, it's crucial first to rearrange the formula to solve for the unknown variable. In our exercise, we need to find the new pressure \( P_2 \). Start with the equation:\[ P_2 = P_1 \cdot \frac{V_1}{V_2} \cdot \frac{T_2}{T_1} \] Let's break down the steps:
  • Rearrange: Solve the equation for \( P_2 \) if it's not already isolated.
  • Substitute Values: Plug in the given values into the equation.
  • Calculate: Perform the multiplication and division necessary to solve for \( P_2 \).
In the exercise, the calculation showed us that compressing and cooling the gas from the initial conditions resulted in a final pressure of approximately 4.35 atm.
Initial and Final Conditions
Understanding the initial and final conditions of a gas is essential in applying the Combined Gas Law effectively. These conditions specifically refer to the states of the gas鈥攊ts pressure, volume, and temperature鈥攂efore and after any changes.
  • **Initial Conditions:** These are the starting points for your calculations. In this exercise, the initial pressure \( P_1 \) is 2.00 atm, volume \( V_1 \) is 56.44 L, and temperature \( T_1 \) is 310 K.

  • **Final Conditions:** These values reflect the state of the gas after changes have occurred. For the problem at hand, the final volume \( V_2 \) is 23.52 L, and the final temperature \( T_2 \) is 281 K. The final pressure \( P_2 \) is what we need to calculate.
The ability to identify these conditions clearly ensures the correct application of the Combined Gas Law and accurate determination of unknown gas states.

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Most popular questions from this chapter

(Chemical Connections 5 B) In carbon monoxide poisoning, the hemoglobin is incapable of transporting oxygen to the tissues. How does the oxygen get delivered to the cells when a patient is put into a hyperbaric chamber?

Answer true or false. (a) For a sample of gas at constant temperature, its pressure multiplied by its volume is a constant. (b) For a sample of gas at constant temperature, increasing the pressure increases the volume. (c) For a sample of gas at constant temperature, \(P_{1} / V_{1}=P_{2} / V_{2}\) (d) As a gas expands at constant temperature, its volume increases. (e) The volume of a sample of gas at constant pressure is directly proportional to its temperature-the higher its temperature, the greater its volume. (f) A hot-air balloon rises because hot air is less dense than cooler air. (g) For a gas sample in a container of fixed volume, an increase in temperature results in an increase in pressure. (h) For a gas sample in a container of fixed volume, \(P \times T\) is a constant. (i) When steam at \(100^{\circ} \mathrm{C}\) in an autoclave is heated to \(120^{\circ} \mathrm{C}\), the pressure within the autoclave increases. (j) When a gas sample in a flexible container at constant pressure at \(25^{\circ} \mathrm{C}\) is heated to \(50^{\circ} \mathrm{C}\), its volume doubles. (k) Lowering the diaphragm causes the chest cavity to increase in volume and the pressure of air in the lungs to decrease. (l) Raising the diaphragm decreases the volume of the chest cavity and forces air out of the lungs.

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(Chemical Connections 5 D) Why is the damage by severe frostbite irreversible?
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