Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 50

An ideal gas that occupies \(1.2 \mathrm{m}^{3}\) at a pressure of $1.0 \times 10^{5} \mathrm{Pa}\( and a temperature of \)27^{\circ} \mathrm{C}$ is compressed to a volume of \(0.60 \mathrm{m}^{3}\) and heated to a temperature of \(227^{\circ} \mathrm{C} .\) What is the new pressure?

Short Answer

Expert verified
Answer: The new pressure after the compression and heating process is approximately \(2.0\times10^{5}\,\mathrm{Pa}\).

Step by step solution

01

Write the Ideal Gas Law for the initial and final states

We are given information on the initial and final states of the ideal gas: initial volume (V1), initial pressure (P1), initial temperature (T1), final volume (V2), and final temperature (T2). We can write the Ideal Gas Law for these two states as: $$ P_{1}V_{1}=nRT_{1} \quad \text{and} \quad P_{2}V_{2}=nRT_{2} $$ Since the number of gas particles (n) and the gas constant (R) don't change, we can equate the two expressions.
02

Equate and simplify the Ideal Gas Law expressions for the initial and final states

We can now equate the expressions for the initial and final states and solve for the new pressure, P2: $$ \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}} $$
03

Plug in the given values and convert temperatures to Kelvin

Before we plug in the given values, make sure to convert the temperatures from Celsius to Kelvin by adding 273.15: $$ T_{1}=27^{\circ}\mathrm{C}+273.15=300.15\,\mathrm{K} \\ T_{2}=227^{\circ}\mathrm{C}+273.15=500.15\,\mathrm{K} $$ Now insert the given values into the equation: $$ \frac{1.0\times10^{5}\mathrm{Pa}\cdot1.2\mathrm{m}^{3}}{300.15\,\mathrm{K}}=\frac{P_{2}(0.6\mathrm{m}^{3})}{500.15\,\mathrm{K}} $$
04

Solve for the new pressure, P2

To find the new pressure, P2, we can now solve the equation: $$ P_{2}=\frac{1.0\times10^{5}\mathrm{Pa}\cdot1.2\mathrm{m}^{3}\cdot500.15\,\mathrm{K}}{300.15\,\mathrm{K}\cdot0.6\mathrm{m}^{3}} $$ Calculating the result gives us: $$ P_{2}\approx2.0\times10^{5}\,\mathrm{Pa} $$ The new pressure after the compression and heating process is approximately \(2.0\times10^{5}\,\mathrm{Pa}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Find the molar mass of ammonia (NH \(_{3}\) ).
Steel railroad tracks of length \(18.30 \mathrm{m}\) are laid at $10.0^{\circ} \mathrm{C} .$ How much space should be left between the track sections if they are to just touch when the temperature is \(50.0^{\circ} \mathrm{C} ?\)
How many hydrogen atoms are present in \(684.6 \mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) ?\)
For divers going to great depths, the composition of the air in the tank must be modified. The ideal composition is to have approximately the same number of \(\mathrm{O}_{2}\) molecules per unit volume as in surface air (to avoid oxygen poisoning), and to use helium instead of nitrogen for the remainder of the gas (to avoid nitrogen narcosis, which results from nitrogen dissolving in the bloodstream). Of the molecules in dry surface air, \(78 \%\) are \(\mathrm{N}_{2}, 21 \%\) are \(\mathrm{O}_{2},\) and \(1 \%\) are \(\mathrm{Ar}\). (a) How many \(\mathrm{O}_{2}\) molecules per \(\mathrm{m}^{3}\) are there in surface air at \(20.0^{\circ} \mathrm{C}\) and 1.00 atm? (b) For a diver going to a depth of \(100.0 \mathrm{m}\) what percentage of the gas molecules in the tank should be \(\mathrm{O}_{2} ?\) (Assume that the density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\) and the temperature is $20.0^{\circ} \mathrm{C} .$ )
(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Electricity and Magnetism

Read Explanation

Solid State Physics

Read Explanation

Mechanics and Materials

Read Explanation

Thermodynamics

Read Explanation

Energy Physics

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.

Sign-up for free