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Chapter 14: Problem 62

Helium (He), a monatomic gas, fills a 0.010-m \(^{3}\) container. The pressure of the gas is \(6.2 \times 10^{5}\) Pa. How long would a 0.25 -hp engine have to run \((1 \mathrm{hp}=746 \mathrm{W})\) to produce an amount of energy equal to the internal energy of this gas?

Short Answer

Expert verified
The engine must run for approximately 49.91 seconds to equal the helium's internal energy.

Step by step solution

01

Calculate Internal Energy per Mole

For a monatomic ideal gas like helium, the molar internal energy is given by \( U = \frac{3}{2} nRT \) where \( n \) is the number of moles, \( R \) is the ideal gas constant \( 8.314 \, \text{J/mol K} \), and \( T \) is the temperature in Kelvin. However, here we use \( U = \frac{3}{2} PV \) because we are given pressure and volume. The internal energy for the gas is therefore \( U = \frac{3}{2} \times 6.2 \times 10^5 \, \text{Pa} \times 0.010 \, \text{m}^3 \).
02

Calculate Total Internal Energy

Substitute the given values into the equation: \( U = \frac{3}{2} \times 6.2 \times 10^5 \, \text{Pa} \times 0.010 \, \text{m}^3 = 9310 \, \text{J} \). Thus, the internal energy of the gas is 9310 Joules.
03

Convert Horsepower to Watts

We need to calculate energy in terms of power to find the time it takes to produce the internal energy using a 0.25-hp engine. First, convert horsepower to watts: \( 0.25 \times 746 \, \text{W/hp} = 186.5 \, \text{W} \).
04

Calculate Time Required

The energy output of the engine is \( Power \times Time \). Rearrange the formula to calculate time: \( Time = \frac{Energy}{Power} \). For \( U = 9310 \, \text{J} \) and \( Power = 186.5 \, \text{W} \), we find \( Time = \frac{9310}{186.5} \approx 49.91 \, \text{seconds} \).

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Key Concepts

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

Internal Energy
Internal Energy is a fundamental concept in thermodynamics referring to the total energy contained within a system due to both kinetic and potential energies of particles. In simple terms, it's the energy required to create the system. For gases, especially monatomic gases like helium, the internal energy only depends on the kinetic energy of the particles:
  • This kinetic energy is linked directly to the temperature of the system.
  • For the helium gas, the molar internal energy is expressed as: \( U = \frac{3}{2} nRT \).
  • In our exercise, the Ideal Gas Law simplifies to \( U = \frac{3}{2} PV \) because pressure and volume are given.
This indicates that the higher the pressure or volume, the higher the internal energy, keeping temperature constant.
The internal energy of the helium gas in the problem was calculated to be 9310 Joules using these principles.
Ideal Gas Law
The Ideal Gas Law is an equation that describes the behavior of gases under different conditions. It's a cornerstone in understanding gas properties in chemistry and physics. The formula is:\[ PV = nRT \]where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the ideal gas constant (\(8.314 \, \text{J/mol K}\)), and
  • \( T \) is the temperature in Kelvin.
To connect this to our exercise, the pressure and volume were used to derive the internal energy of helium. This law helps us assume gas behavior is ideal鈥攖hat is, the gas particles do not interact except for elastic collisions. However, in real conditions, deviations can occur, but for typical classroom problems, the Ideal Gas Law is sufficient.
Therefore, understanding the relationship between these variables allows us to predict how changes in one affect the others.
Pressure and Volume Relationship
Pressure and Volume Relationship in the context of gases is often highlighted by Boyle's Law, which states that for a given mass at constant temperature, the pressure of a gas is inversely proportional to its volume:\[ P \propto \frac{1}{V} \]However, when both pressure and volume are considered at a constant temperature or when used in the Ideal Gas Law, they highlight the idea that as volume increases, more space for gas particles results in fewer collisions, thus less pressure, if temperature is unchanged.
  • In simple terminology, if you squeeze a gas into a smaller space, the pressure it exerts becomes higher.
  • Similarly, reducing the pressure (assuming no temperature change) allows the gas to expand and occupy more volume.
  • In our exercise, volume and pressure were given specific values, showing how these quantities help determine the internal energy of the helium gas.
This relationship is crucial for solving many thermodynamic problems where the behavior of gases is involved.

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Most popular questions from this chapter

A gas fills the right portion of a horizontal cylinder whose radius is \(5.00 \mathrm{cm} .\) The initial pressure of the gas is \(1.01 \times 10^{5} \mathrm{Pa}\) A frictionless movable piston separates the gas from the left portion of the cylinder, which is evacuated and contains an ideal spring, as the drawing shows. The piston is initially held in place by a pin. The spring is initially unstrained, and the length of the gas-filled portion is \(20.0 \mathrm{cm} .\) When the pin is removed and the gas is allowed to expand, the length of the gas-filled chamber doubles. The initial and final temperatures are equal. Determine the spring constant of the spring.

When you push down on the handle of a bicycle pump, a piston in the pump cylinder compresses the air inside the cylinder. When the pressure in the cylinder is greater than the pressure inside the inner tube to which the pump is attached, air begins to flow from the pump to the inner tube. As a biker slowly begins to push down the handle of a bicycle pump, the pressure inside the cylinder is \(1.0 \times 10^{5} \mathrm{Pa}\), and the piston in the pump is \(0.55 \mathrm{m}\) above the bottom of the cylinder. The pressure inside the inner tube is \(2.4 \times 10^{5}\) Pa. How far down must the biker push the handle before air begins to flow from the pump to the inner tube? Ignore the air in the hose connecting the pump to the inner tube, and assume that the temperature of the air in the pump cylinder does not change.

The mass of a hot-air balloon and its occupants is \(320 \mathrm{kg}\) (excluding the hot air inside the balloon). The air outside the balloon has a pressure of \(1.01 \times 10^{5}\) Pa and a density of \(1.29 \mathrm{kg} / \mathrm{m}^{3}\). To lift off, the air inside the balloon is heated. The volume of the heated balloon is \(650 \mathrm{m}^{3} .\) The pressure of the heated air remains the same as the pressure of the outside air. To what temperature (in kelvins) must the air be heated so that the balloon just lifts off? The molecular mass of air is 29 u.

A container holds 2.0 mol of gas. The total average kinetic energy of the gas molecules in the container is equal to the kinetic energy of an \(8.0 \times 10^{-3}-\mathrm{kg}\) bullet with a speed of \(770 \mathrm{m} / \mathrm{s}\). What is the Kelvin temperature of the gas?

A tube has a length of \(0.015 \mathrm{m}\) and a cross-sectional area of \(7.0 \times \mathrm{x}\) \(10^{-4} \mathrm{m}^{2} .\) The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is \(5.0 \times 10^{-10} \mathrm{m}^{2} / \mathrm{s} .\) A difference in concentration of \(3.0 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{3}\) is maintained between the ends of the tube. How much time is required for \(8.0 \times 10^{-13} \mathrm{kg}\) of sucrose to be transported through the tube?
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