Thermodynamic problem using ideal gas( me (T_T) )

In summary: KPA(gage). He starts with 200 KPA(gage) and 23'C in the tires, and after a long drive at 83'C and with 0.11kg of air in each tire, the pressure is 182.48 KPA(gage) and 0.0064kg of air has escaped. In summary, the motorist used the formula PV=mRT and derived the equations m1 = P1V1/R(T1) and m2 = P2V2/R(T2) to determine the mass of air escaping from each tire and the pressure of the tire when the temperature returns to
  • #1
jabjab
1
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
 
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  • #2
Isn't the equation PV=nRT where n is the number of moles of the gas?
 
  • #3
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
 

1. What is an ideal gas in thermodynamics?

An ideal gas is a theoretical gas that follows the ideal gas law, which states that the pressure, volume, and temperature of the gas are all directly proportional to each other. In other words, as one variable increases, the others also increase in proportion. Ideal gases have no intermolecular forces and their molecules have negligible volume.

2. What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. It is represented by the equation PV = nRT, where R is the gas constant.

3. How do you solve a thermodynamic problem using an ideal gas?

To solve a thermodynamic problem using an ideal gas, you need to first identify the known variables (P, V, n, and T) and the unknown variable. Then, you can use the ideal gas law to calculate the unknown variable. It is important to make sure that all units are consistent and to convert them if necessary.

4. What are the units of the ideal gas law?

The units of the ideal gas law depend on the units used for pressure, volume, number of moles, and temperature. In SI units, the units of the ideal gas law are Pascals (Pa) for pressure, cubic meters (m^3) for volume, moles (mol) for number of moles, and Kelvin (K) for temperature. In other unit systems, such as the Imperial system, different units may be used.

5. What are some real-life applications of thermodynamics using ideal gases?

Thermodynamics using ideal gases has many practical applications in various fields, including chemistry, physics, and engineering. Some examples include studying the behavior of gases in chemical reactions, designing and optimizing engines and power plants, and understanding the properties of air and other gases in the Earth's atmosphere. It is also used in the development of new materials and technologies, such as refrigeration systems and fuel cells.

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