/*! 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 48 A sample of hydrogen exerts a pr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 48

A sample of hydrogen exerts a pressure of 0.329 atm at \(47^{\circ} \mathrm{C}\) . What will the pressure bo at \(77^{\circ} \mathrm{C},\) assuming constant volume?

Short Answer

Expert verified
The pressure of the hydrogen at \(77^{\circ} \mathrm{C}\), assuming constant volume, will be approximately \(0.361 atm\).

Step by step solution

01

Convert temperatures to Kelvin

First, convert the temperatures from Celsius to Kelvin. The Kelvin temperature is the Celsius temperature plus 273.15. Thus, \(T1 = 47^{\circ} \mathrm{C} + 273.15 = 320.15 K\) and \(T2 = 77^{\circ} \mathrm{C} + 273.15 = 350.15 K\).
02

Apply pressure-temperature relationship

According to the law of gases, at a constant volume, the pressure of a gas is directly proportional to its temperature. That is \(\frac{P1}{T1} = \frac{P2}{T2}\). Where, \(P1 = 0.329 atm\) and \(T1 = 320.15 K\) are the initial pressure and temperature, and \(T2 = 350.15 K\) is the final temperature. The task is to solve for \(P2\), the pressure at \(77^{\circ} \mathrm{C}\).
03

Solve for the final pressure P2

Rearrange the equation to solve for the final pressure: \(P2 = P1 \times \frac{T2}{T1}\) = \(0.329 atm \times \frac{350.15 K}{320.15 K}\) = \(0.361 atm\)

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!

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

An unknown gas effuses at a speed one- quarter of that of helium. What is the molar mass of the unknown gas? It is either sulfur dioxide or sulfur trioxide. Which gas is it?

Write a paragraph that describes how the kinetic-molecular theory explains the following properties of a gas: fluidity, compressibility, and pressure.

What relationship does Gay-Lussac's law express?

What relationship does Charles's law express?

How do pascals relate to newtons?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

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.