# Solving Gauge Pressure: Automobile Tire at 20C and 50C

• ch3570r
In summary, the conversation discusses finding the gauge pressure of air in an automobile tire after its temperature increases from 20C to 50C. Using the equation (p2v2/t2)=(p1V1/t1), with known values of 200kPa for initial pressure, 1 for constant volume, and 293K and 323K for initial and final temperatures, respectively, the summary concludes that the gauge pressure of the air in the tire should be 220 kPa instead of the expected 230 kPa. It is noted that this discrepancy may be due to the comment in the question about gauge pressure.
ch3570r
"An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is 20C. (Gauge pressure is the difference between the actual pressure and atmostpheric pressure.) After the car has been driven at high speeds, the tires temperature increases to 50C. Assuming that the volume of the tire does not change, and that air behaves as an ideal gas, find the gauge pressure of the air in the tire."

I will use (p2v2/t2)=(p1V1/t1)

(pressure one) p1 = 200kPa
(pressure two) p2 = ?
(volume one) v1 = 1 (because its a constant)
(volume two) v2 = 1 (again, constant)
(temperature one) t1 = 20C = 293 K
(temperature two) t2 = 50K = 323 K

I solve for p2, but i get 220, not 230 as the book says. It might have something to due with the comment in the question about gauge pressure. Anyone see the problem?

ch3570r said:
"An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is 20C. (Gauge pressure is the difference between the actual pressure and atmostpheric pressure.) After the car has been driven at high speeds, the tires temperature increases to 50C. Assuming that the volume of the tire does not change, and that air behaves as an ideal gas, find the gauge pressure of the air in the tire."

I will use (p2v2/t2)=(p1V1/t1)

(pressure one) p1 = 200kPa
(pressure two) p2 = ?
(volume one) v1 = 1 (because its a constant)
(volume two) v2 = 1 (again, constant)
(temperature one) t1 = 20C = 293 K
(temperature two) t2 = 50K = 323 K

I solve for p2, but i get 220, not 230 as the book says. It might have something to due with the comment in the question about gauge pressure. Anyone see the problem?

You're correct in that you must use the actual pressure, not the gauge pressure. Re-try your calculation.

I would first like to clarify that the term "gauge pressure" refers to the pressure measured by a gauge, which is the difference between the actual pressure and atmospheric pressure. In this case, the given gauge pressure of 200 kPa is already the difference between the actual pressure and atmospheric pressure. Therefore, we do not need to adjust for atmospheric pressure in our calculations.

Using the ideal gas law, we can solve for the new gauge pressure (p2) using the given values:

(p1V1)/t1 = (p2V2)/t2

Substituting our values:

(200 kPa)(1 L)/(293 K) = (p2)(1 L)/(323 K)

Solving for p2, we get:

p2 = (200 kPa)(323 K)/(293 K) = 220 kPa

Therefore, the gauge pressure of the air in the tire at 50C is 220 kPa, not 230 kPa as stated in the question. It is possible that the book made a mistake or used a different method to calculate the gauge pressure. As scientists, it is important to always double check our calculations and results, and to also consider potential sources of error.

## 1. How do I calculate gauge pressure for an automobile tire at different temperatures?

To calculate gauge pressure for an automobile tire at different temperatures, you will need to use the ideal gas law equation: P = nRT/V, where P is pressure, n is the number of moles, R is the gas constant, T is temperature, and V is volume. First, determine the number of moles of gas in the tire by dividing the mass of the air in the tire by the molar mass of air. Then, plug in the values for temperature, volume, and the gas constant (for air, use 0.0821 L·atm/mol·K) to calculate the gauge pressure at the desired temperature.

## 2. How does temperature affect the gauge pressure of an automobile tire?

As temperature increases, the gauge pressure of an automobile tire will also increase. This is because the molecules of gas inside the tire have more energy and are moving faster, causing them to exert a greater force on the walls of the tire, resulting in a higher pressure. Conversely, as temperature decreases, the gauge pressure will decrease.

## 3. Can I use the ideal gas law to calculate gauge pressure for any type of gas?

Yes, the ideal gas law can be used to calculate gauge pressure for any type of gas as long as the gas behaves like an ideal gas. This means that the gas molecules are far apart and do not interact with each other, and there are no significant intermolecular forces present. Real gases may deviate from ideal gas behavior, especially at high pressures and low temperatures.

## 4. What is the difference between gauge pressure and absolute pressure?

Gauge pressure is the pressure measured relative to atmospheric pressure, while absolute pressure is the total pressure exerted by a gas, including atmospheric pressure. Gauge pressure is often used for practical purposes, such as measuring tire pressure, while absolute pressure is used in scientific calculations.

## 5. How does gauge pressure affect the performance of a car tire?

Gauge pressure is a critical factor in determining the performance of a car tire. If the gauge pressure is too low, the tire will not have enough air to support the weight of the vehicle and can result in poor handling, increased wear, and potential tire failure. On the other hand, if the gauge pressure is too high, it can result in a harsher ride, decreased traction, and uneven tire wear. It is essential to regularly check and maintain the proper gauge pressure for optimal tire performance.

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