Thermodynamic problem using ideal gas( me (T_T) )

Click For Summary
SUMMARY

The discussion revolves around solving a thermodynamic problem involving an automobile tire's air pressure and temperature changes. The motorist starts with a tire pressure of 200 KPA(gage) at 23°C, which increases to 83°C during a trip. The correct answers derived from the calculations are a mass of air escaping each tire of 0.0064 kg and a tire pressure of 182.48 KPA(gage) when the temperature returns to 23°C. The forum participants emphasize the importance of using the ideal gas law, PV=nRT, and calculating the number of moles of air to solve the problem accurately.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of thermodynamic principles, specifically relating to pressure and temperature
  • Familiarity with gas constants, particularly R = 8.314 Joules/mole K
  • Basic algebra for manipulating equations and solving for variables
NEXT STEPS
  • Learn how to apply the ideal gas law in various thermodynamic scenarios
  • Research the relationship between pressure, volume, and temperature in gases
  • Study the concept of moles and how to convert mass to moles for gases
  • Explore real-world applications of thermodynamics in automotive engineering
USEFUL FOR

Students studying thermodynamics, automotive engineers, and anyone interested in the practical applications of the ideal gas law in real-world scenarios.

jabjab
Messages
1
Reaction score
0

Homework Statement


hi,
can anyone help me with this problem?

a motorist equips his automobile tire with a relief type valve so that the pressure inside the tire never exceeds 240 KPA(gage). He starts a trip with a pressure of 200 KPA(gage) and a temp. 23'C in the tires.During the long drive the temp. of air in the tire reaches 83'C each tire contains 0.11kg of air.a) the mass of air escaping each tire.b) the pressure of the tire when temp. return to 23'C

given the answer provide the solution:

answer a)0.0064kg b)182.48 KPA(gage)



Homework Equations





The Attempt at a Solution


for a)
i use the formula PV=mRT and derive equation:

m1=P1V1/R(T1)
m2=P2V2/R(T2)

i assume volume is constant because on the problem volume is not given
m1= ((200KJ/m3)(1m3)(1000J/1KJ))/((286.9J/Kg.K)(296K)
=20000/84922.4
=2.35509kg

m2=((240)(1)(1000))/((286.9)(356))
= 240000/102136.4
= 234980kg

m1-m2=m
m= 2.35509kg-2.34980kg
m= 0.00529kg


for b)
i use the formula
P1/T1=P2/T2

P2=P1(T2/T1)
=240KPA(gage)(296/356)
=199 KPA(gage)

my answer doesn't match the answer provided in the problem.can you please help me...T_T
 
Physics news on Phys.org
Isn't the equation PV=nRT where n is the number of moles of the gas?
 
You have to determine the number of moles of air in the tire from the mass of air. Then you have to find the volume of the tire using PV = nRT where R is the gas constant 8.314 Joules/mole K.

What is the number of moles of air in the tire to begin at T = 296 K? Find V (which is the same at all times).

Then just apply PV=nRT.

AM
 

Similar threads

Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
7
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
9K
Replies
9
Views
4K