A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a constant pressure of 1.35 x 105 Pa. The box is placed over a fire, causing the volume to increase from 0.00200 m³ to 0.00300 m³. Find the increase in temperature of the gas. (For N2 molar mass M = 28 grams.)

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter16: Temperature And The Kinetic Theory Of Gases
Section: Chapter Questions
Problem 38P
icon
Related questions
Question
A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a
constant pressure of 1.35 x 10$ Pa. The box is placed over a fire, causing the
volume to increase from 0.00200 m3 to 0.00300 m³. Find the increase in
temperature of the gas.
(For N2 molar mass M = 28 grams.)
Transcribed Image Text:A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a constant pressure of 1.35 x 10$ Pa. The box is placed over a fire, causing the volume to increase from 0.00200 m3 to 0.00300 m³. Find the increase in temperature of the gas. (For N2 molar mass M = 28 grams.)
Formulas:
Work done on a gas: Wgas
= - PAV
First Law of Thermodynamics: AU = Q+ W
Idea gas law: PV = nRT, thus, PAV = NRAT
Molar specific heat at constant volume: Cv : AU = nCvAt where n is moles
3
R for monatomic gasses,
2
5
R for diatomic gasses
2
The value of Cv is just:
5
R for monatomic gasses,
2
7
for diatomic gasses
2
The value of Cp is just:
Joules
Where R is the universal gas constant: R = 8.31
mole-Kelvin
Molar specific heat at constant pressure: Q = CPAT
CP
For adiabatic processes:
PVY = constant, where
Y
Cv
Work done at constant temperature:
W = -nRT In(
1
The First Law and Thermodynamic Processes (Ideal Gases)
Process
AU
Q
W
Isobaric
nC, AT
nC, AT
-P AV
Adiabatic
nC, AT
Δυ
Isovolumetric
nC, AT
AU
Isothermal
-W
-nRT In
Transcribed Image Text:Formulas: Work done on a gas: Wgas = - PAV First Law of Thermodynamics: AU = Q+ W Idea gas law: PV = nRT, thus, PAV = NRAT Molar specific heat at constant volume: Cv : AU = nCvAt where n is moles 3 R for monatomic gasses, 2 5 R for diatomic gasses 2 The value of Cv is just: 5 R for monatomic gasses, 2 7 for diatomic gasses 2 The value of Cp is just: Joules Where R is the universal gas constant: R = 8.31 mole-Kelvin Molar specific heat at constant pressure: Q = CPAT CP For adiabatic processes: PVY = constant, where Y Cv Work done at constant temperature: W = -nRT In( 1 The First Law and Thermodynamic Processes (Ideal Gases) Process AU Q W Isobaric nC, AT nC, AT -P AV Adiabatic nC, AT Δυ Isovolumetric nC, AT AU Isothermal -W -nRT In
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Kinetic theory of gas
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
University Physics Volume 2
University Physics Volume 2
Physics
ISBN:
9781938168161
Author:
OpenStax
Publisher:
OpenStax
An Introduction to Physical Science
An Introduction to Physical Science
Physics
ISBN:
9781305079137
Author:
James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:
Cengage Learning