/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A monatomic ideal gas is heated ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 42

A monatomic ideal gas is heated while at a constant volume of \(1.00 \times 10^{-3} \mathrm{m}^{3},\) using a ten-watt heater. The pressure of the gas increases by \(5.0 \times 10^{4}\) Pa. How long was the heater on?

Short Answer

Expert verified
The heater was on for 7.5 seconds.

Step by step solution

01

Understand the Problem

We are asked to determine how long a ten-watt heater was used to heat a monatomic ideal gas at constant volume, given that the pressure increases by \(5.0 \times 10^{4}\) Pa.
02

Recall the Ideal Gas Law

The ideal gas law is given by \( PV = nRT \), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature (in Kelvin). Since volume is constant, an increase in pressure implies an increase in temperature.
03

Relate Change in Pressure to Change in Temperature

For a fixed volume, the change in pressure \(\Delta P\) is proportional to the change in temperature \(\Delta T\) by the formula: \( \Delta P \cdot V = nR\Delta T \). Use this to express the change in temperature.
04

Calculate Change in Temperature

Given that \(\Delta P = 5.0 \times 10^{4} \) Pa and \( V = 1.00 \times 10^{-3} \) m鲁,\( nR\Delta T = \Delta P \cdot V = (5.0 \times 10^{4}) \cdot (1.00 \times 10^{-3}) = 50 \).
05

Use the First Law of Thermodynamics

The first law states \( Q = \Delta U \) for constant volume, where \( Q \) is the heat added and \( \Delta U = \frac{3}{2} n R \Delta T \) for a monatomic ideal gas. Thus, \( Q = \frac{3}{2} \times 50 = 75 \mathrm{J} \).
06

Relate Power and Time to Heat Added

Power is defined as \( P = \frac{Q}{t} \), where \( P = 10 \) W (watts) and \( Q = 75 \) Joules. Solve for time \( t \): \( t = \frac{Q}{P} = \frac{75}{10} = 7.5 \mathrm{s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Monatomic Gas
Monatomic gases are the simplest form of gases, consisting of single atoms. Typical examples include noble gases such as helium or neon. Because they have no rotational or vibrational modes, they simplify thermodynamic calculations. In a monatomic ideal gas, energy changes primarily affect the translational kinetic energy of the atoms.
This means energy calculations for monatomic gases are straightforward. The internal energy (U) of a monatomic ideal gas is given by:
  • \( U = \frac{3}{2} nRT \)
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. This relationship is crucial when applying the first law of thermodynamics. Using these principles helps determine how heat inputs, like the one from a heater, affect the gas.
Pressure Change
Pressure change in a gas refers to the variation in pressure resulting from changes in temperature, volume, or the amount of gas. According to the ideal gas law, \( PV = nRT \), pressure \( P \) increases with temperature \( T \) if volume \( V \) is held constant.
For the problem of heating a monatomic gas at constant volume, an increase in pressure directly relates to a rise in temperature. This is because, with constant volume, the ideal gas law simplifies to show that pressure is directly proportional to temperature:
  • \( P \propto T \)
The given exercise involves an increase in pressure, \( \Delta P = 5.0 \times 10^{4} \) Pa, calculated by knowing the heat added.
First Law of Thermodynamics
The first law of thermodynamics is a form of the energy conservation principle. It states that the total energy of a system and its surroundings remains constant:
The first law is written as:
  • \( Q = \Delta U + W \)
where \( Q \) is the heat added to the system, \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the system. For processes at constant volume, such as in the exercise, there is no work done by the gas \((W = 0)\), thus simplifying to \( Q = \Delta U \).
For a monatomic ideal gas, the change in internal energy \( \Delta U \) is
  • \( \Delta U = \frac{3}{2} n R \Delta T \)
This provides an essential step to solve how long it took for the heater to bring about a change in the pressure of the gas.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Each of two Carnot engines uses the same cold reservoir at a temperature of \(275 \mathrm{K}\) for its exhaust heat. Each engine receives \(1450 \mathrm{J}\) of input heat. The work from either of these engines is used to drive a pulley arrangement that uses a rope to accelerate a \(125-\mathrm{kg}\) crate from rest along a horizontal frictionless surface, as shown in the figure. With engine 1 the crate attains a speed of \(2.00 \mathrm{m} / \mathrm{s},\) while with engine 2 it attains a speed of \(3.00 \mathrm{m} / \mathrm{s} .\) Concepts: (i) With which engine is the change in the crate's energy greater? (ii) Which engine does more work? Explain your answer. (iii) For which engine is the temperature of the hot reservoir greater? \(\mathrm{Cal}\) culations: Find the temperature of the hot reservoir for each engine.

The temperatures indoors and outdoors are 299 and \(312 \mathrm{K}\), respectively. A Carnot air conditioner deposits \(6.12 \times 10^{5} \mathrm{J}\) of heat outdoors. How much heat is removed from the house?

Three moles of neon expand isothermally to 0.250 from \(0.100 \mathrm{m}^{3}\). Into the gas flows \(4.75 \times 10^{3} \mathrm{J}\) of heat. Assuming that neon is an ideal gas, find its temperature.

A Carnot engine has an efficiency of \(0.40 .\) The Kelvin temperature of its hot reservoir is quadrupled, and the Kelvin temperature of its cold reservoir is doubled. What is the efficiency that results from these changes?

Three moles of a monatomic ideal gas are heated at a constant volume of \(1.50 \mathrm{m}^{3} .\) The amount of heat added is \(5.24 \times 10^{3} \mathrm{J} .\) (a) What is the change in the temperature of the gas? (b) Find the change in its internal energy. (c) Determine the change in pressure.
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Rotational Dynamics

Read Explanation

Dynamics

Read Explanation

Solid State Physics

Read Explanation

Mechanics and Materials

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.