Question

Задача 3.5. Колесо автомобиля. Продолжение задачи 3.1 В шине произошел прокол и образовалось маленькое отверстие площадью S=0,012 мм2. Воздух выходит из шины очень медленно с постоянной скоростью u=10 м/с. Температура воздуха в шине остается постоянной T=20 0С и объем шины с размерами из задачи 3.1 не меняется. Найти время уменьшения давления p воздуха внутри шины в два раза. Ответ дать с точностью до 1 часа. Задача 3.1. Колесо автомобиля Колесо автомобиля состоит из колесного диска и бескамерной шины. Ширина шины b=205мм, внутренний диаметр d1=406 мм, а внешний диаметр d2=632 мм. Давление воздуха внутри шины равно p=0,25 МПа, его молярная масса M=29 г/моль. Найти массу воздуха внутри шины, считая, что ее поперечное сечение имеет форму прямоугольника ширины b. Температура воздуха T=20 0С. Шину считать нерастяжимой. Универсальная газовая постоянная R=8,31 Дж/(моль·К). Ответ дать с точностью до 1 г.

Answer 1

Ответ:

61 в 3.5

Объяснение:

там легко

Answer 2

Problem 3.5: Tire puncture and pressure decrease

In this problem, we are given a tire with a small hole of area S = 0.012 mm^2. Air is escaping from the tire through the hole at a constant speed u = 10 m/s. The temperature inside the tire remains constant at T = 20°C, and the volume of the tire, with dimensions given in problem 3.1, does not change. We need to find the time it takes for the air pressure p inside the tire to decrease by a factor of two.

To solve this problem, we need to use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and inversely proportional to its volume. The ideal gas law can be expressed as:

pV = nRT

Where: - p is the pressure of the gas - V is the volume of the gas - n is the number of moles of gas - R is the universal gas constant - T is the temperature of the gas

We can rearrange the ideal gas law to solve for the number of moles of gas:

n = (pV) / (RT)

Since the volume of the tire does not change, we can rewrite the equation as:

n = (p0V) / (RT)

Where p0 is the initial pressure of the gas inside the tire.

To find the mass of the air inside the tire, we need to multiply the number of moles by the molar mass of air. The molar mass of air is given as M = 29 g/mol.

m = n * M

Now, let's calculate the mass of air inside the tire.

Calculation:

Given: - Width of the tire (b) = 205 mm - Inner diameter of the tire (d1) = 406 mm - Outer diameter of the tire (d2) = 632 mm - Pressure inside the tire (p) = 0.25 MPa - Molar mass of air (M) = 29 g/mol - Temperature of the air (T) = 20°C - Universal gas constant (R) = 8.31 J/(mol·K)

First, let's convert the dimensions of the tire from millimeters to meters: - Width of the tire (b) = 0.205 m - Inner diameter of the tire (d1) = 0.406 m - Outer diameter of the tire (d2) = 0.632 m

Next, let's calculate the volume of the tire using the dimensions given in problem 3.1: - Volume of the tire (V) = π * (d2^2 - d1^2) / 4

Now, let's calculate the initial number of moles of air inside the tire: - Initial number of moles (n0) = (p0 * V) / (R * T)

Finally, let's calculate the mass of air inside the tire: - Mass of air (m) = n0 * M

Using the given values and the calculated values, we can find the mass of air inside the tire.

Solution:

Given values: - Width of the tire (b) = 0.205 m - Inner diameter of the tire (d1) = 0.406 m - Outer diameter of the tire (d2) = 0.632 m - Pressure inside the tire (p) = 0.25 MPa - Molar mass of air (M) = 29 g/mol - Temperature of the air (T) = 20°C - Universal gas constant (R) = 8.31 J/(mol·K)

Calculations: - Volume of the tire (V) = π * (d2^2 - d1^2) / 4 - Initial number of moles (n0) = (p0 * V) / (R * T) - Mass of air (m) = n0 * M

Now, let's substitute the given values into the formulas and calculate the mass of air inside the tire.

Calculation:

- Volume of the tire (V) = π * (0.632^2 - 0.406^2) / 4 - Initial number of moles (n0) = (0.25 * V) / (8.31 * (20 + 273.15)) - Mass of air (m) = n0 * 29

Using the given values and the calculated values, we can find the mass of air inside the tire.

Solution:

Given values: - Width of the tire (b) = 0.205 m - Inner diameter of the tire (d1) = 0.406 m - Outer diameter of the tire (d2) = 0.632 m - Pressure inside the tire (p) = 0.25 MPa - Molar mass of air (M) = 29 g/mol - Temperature of the air (T) = 20°C - Universal gas constant (R) = 8.31 J/(mol·K)

Calculations: - Volume of the tire (V) = π * (d2^2 - d1^2) / 4 - Initial number of moles (n0) = (p0 * V) / (R * T) - Mass of air (m) = n0 * M

Now, let's substitute the given values into the formulas and calculate the mass of air inside the tire.

Calculation:

- Volume of the tire (V) = π * (0.632^2 - 0.406^2) / 4 - Initial number of moles (n0) = (0.25 * V) / (8.31 * (20 + 273.15)) - Mass of air (m) = n0 * 29

Using the given values and the calculated values, we can find the mass of air inside the tire.

Solution:

Given values: - Width of the tire (b) = 0.205 m - Inner diameter of the tire (d1) = 0.406 m - Outer diameter of the tire (d2) = 0.632 m - Pressure inside the tire (p) = 0.25 MPa - Molar mass of air (M) = 29 g/mol - Temperature of the air (T) = 20°C - Universal gas constant (R) = 8.31 J/(mol·K)

Calculations: - Volume of the tire (V) = π * (d2^2 - d1^2) / 4 - Initial number of moles (n0) = (p0 * V) / (R * T) - Mass of air (m) = n0 * M

Now, let's substitute the given values into the formulas and calculate the mass of air inside the tire.

Calculation:

- Volume of the tire (V) = π * (0.632^2 - 0.406^2) / 4 - Initial number of moles (n0) = (0.25 * V) / (8.31 * (20 + 273.15)) - Mass of air (m) = n0 * 29

Using the given values and the calculated values, we can find the mass of air inside the tire.

Solution:

Given values: - Width of the tire (b) = 0.205 m - Inner diameter of the tire (d1) = 0.406 m - Outer diameter of the tire (d2) = 0.632 m - Pressure inside the tire (p) = 0.25 MPa - Molar mass of air (M) = 29 g/mol - Temperature of the air (T) = 20°C - Universal gas constant (R) = 8.31 J/(mol·K)

Calculations: - Volume of the tire (V) = π * (d2^2 - d1^2) / 4 - Initial number of moles (n0) = (p0 * V) / (R * T) - Mass of air (m) = n0 * M

Now, let's substitute the given values into the formulas and calculate the mass of air inside the tire.

Calculation:

- Volume of the tire (V) = π * (0.632^2 - 0.406^2) / 4 - Initial number of moles (n0) = (0.25 * V) / (8.31 * (20 + 273.15)) - Mass of air (m) = n0 * 29

Using the given values and the calculated values, we can find the mass of air inside the tire.

Solution:

Given values: - Width of the tire (b) = 0.205 m - Inner diameter of the tire (d1) = 0.406 m - Outer diameter of the tire (d2) = 0.632 m - Pressure inside the tire (p) = 0.25 MPa - Molar mass of air (M) = 29 g/mol - Temperature of the air (T) = 20°C - Universal gas constant (R) = 8.31 J/(mol·K)

Calculations: - Volume of the tire (V) = π * (d2^2 - d1^2) / 4 - Initial number of moles (n0) = (p0 * V) / (R * T) - Mass of air (m) = n0