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Methane gas is compressed from \(20 .\) L to \(2.5 \mathrm{L}\) at a constant temperature. The final pressure is 12.2 atm. What was the original pressure?

Short Answer

Expert verified
The original pressure was 1.525 atm.

Step by step solution

01

Identify the Known Variables

We are given the initial volume \( V_1 = 20 \, \text{L} \), final volume \( V_2 = 2.5 \, \text{L} \), and final pressure \( P_2 = 12.2 \, \text{atm} \). We need to find the initial pressure \( P_1 \).
02

Recall Boyle's Law

Boyle's Law states that the pressure of a gas is inversely proportional to its volume when the temperature remains constant, which is expressed as \( P_1 V_1 = P_2 V_2 \).
03

Rearrange Boyle's Law to Solve for Initial Pressure

Rearrange the equation to find \( P_1 \): \[ P_1 = \frac{P_2 V_2}{V_1} \]
04

Substitute the Known Values into the Equation

Substitute the known values into the rearranged equation: \[ P_1 = \frac{12.2 \, \text{atm} \times 2.5 \, \text{L}}{20 \, \text{L}} \]
05

Calculate the Original Pressure

Carry out the calculation: \( P_1 = \frac{30.5}{20} = 1.525 \, \text{atm} \).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics, relating the pressure, volume, and temperature of an ideal gas. It is stated as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the amount of substance in moles,
  • \( R \) is the ideal gas constant, and
  • \( T \) is the temperature in Kelvin.
The equation shows how changes in one property affect the others, assuming the amount of gas remains constant. For example, if we increase the temperature while keeping the amount of gas and volume constant, the pressure must increase.
In the given exercise, Boyle's Law is highlighted, which is derived from the Ideal Gas Law under constant temperature and number of moles. Here, the product \( PV \) remains constant, showcasing the inverse pressure-volume relationship. This specific scenario is essentially a part of the broader Ideal Gas Law.
Gas Compression
Gas compression is the process of reducing the volume of a gas, which leads to an increase in its pressure, provided the temperature is kept constant. In our exercise, we start with methane gas occupying a volume of \(20\, \text{L}\) and compress it to \(2.5\, \text{L}\).
During compression:
  • The molecules are brought closer together,
  • Collisions between gas molecules and the walls of the container increase,
  • This results in increased pressure according to Boyle's Law.
Gas compression is an important process in various industries. For example, it is used in refrigeration systems, air conditioning, and natural gas transportation. Understanding this concept helps in controlling these processes effectively.
Pressure-Volume Relationship
The pressure-volume relationship highlights how the pressure of a gas is inversely proportional to its volume, as expressed by Boyle's Law \( P_1 V_1 = P_2 V_2 \). Keeping temperature constant, if the volume of a gas decreases, the pressure increases, and vice versa.
Let's break it down to understand:
  • When we squeeze a gas into a smaller space (reducing volume), there's less room for the gas particles to move.
  • This leads to more frequent collisions of the particles with the container walls, thus increasing the pressure.
  • The process is reversible. If you increase the container's volume, the pressure will decrease because the particles have more space to move around.
In our exercise, we used this principle to find the original pressure of methane gas as its volume decreased from \(20 \, \text{L}\) to \(2.5 \, \text{L}\). Thus, demonstrating the direct impact of the pressure-volume relationship in analyzing gas behaviors.

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Most popular questions from this chapter

The oxygen tank in a hospital respiratory unit has a pressure of \(4840 \mathrm{mmHg}\). What is the pressure of the oxygen gas in atmospheres?

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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