An automobile tire has a volume of 1.60 x 10-2 m3 and contains air at a gauge pressure (pressure above atmospheric pressure) of 167 kPa when the temperature is 0.00°C. What is the gauge pressure (in kPa) of the air in the tires when its temperature rises to 27.0°C and its volume increases to 1.69 x 10-2 m3? Assume atmospheric pressure is 1.01 x 105 Pa.

Question

An automobile tire has a volume of 1.60 x 10-2 m3 and contains air at a gauge pressure (pressure above atmospheric pressure) of 167 kPa when the temperature is 0.00°C. What is the gauge pressure (in kPa) of the air in the tires when its temperature rises to 27.0°C and its volume increases to 1.69 x 10-2 m3? Assume atmospheric pressure is 1.01 x 105 Pa.

Answer 1

You have to convert gauge pressure to atmospheric, then work the problem, and in the end, convert back to gauge pressure.

Answer 2

I am filling a room temperature thermos flask with boiling water (about 100 °C). The room temperature is 22 °C. What will the water temperature (nearest °C) be at the end of this class?

Answer 3

To solve this problem, we can use the ideal gas law, which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin

First, we need to convert the given temperature from Celsius to Kelvin:

T1 = 0.00°C + 273.15 = 273.15 K (initial temperature)
T2 = 27.0°C + 273.15 = 300.15 K (final temperature)

Next, we need to determine the initial number of moles of gas in the tire. Since we are not given the molar mass of the gas, we can't directly calculate the number of moles.

However, since the tires are filled with air, which is mostly composed of nitrogen gas (N2), we can assume that the molar mass of the gas is approximately 28.97 g/mol.

We can use the ideal gas law to find the initial number of moles (n1) by rearranging the equation:

n1 = PV1 / (RT1)

Substituting the given values:

P1 = 167 kPa + 1.01 x 10^5 Pa = 1.01 x 10^5 Pa + 167 kPa = 1.01 x 10^5 Pa + 167 x 10^3 Pa = 1.01 x 10^5 Pa + 1.67 x 10^5 Pa = 2.68 x 10^5 Pa
V1 = 1.60 x 10^-2 m^3
R = 8.314 J/(mol·K)
T1 = 273.15 K

n1 = (2.68 x 10^5 Pa) * (1.60 x 10^-2 m^3) / (8.314 J/(mol·K) * 273.15 K)

Now we can find the final pressure (P2) by rearranging the ideal gas law equation:

P2 = n2 * R * T2 / V2

Substituting the given values:

V2 = 1.69 x 10^-2 m^3
R = 8.314 J/(mol·K)
T2 = 300.15 K

P2 = n1 * R * T2 / V2

Finally, we substitute the calculated value of n1 into the equation to find P2:

P2 = (2.68 x 10^5 Pa) * (1.60 x 10^-2 m^3) * (8.314 J/(mol·K) * 273.15 K) / (1.69 x 10^-2 m^3)

Calculating this equation will give you the gauge pressure of the air in the tires when its temperature rises to 27.0°C and its volume increases to 1.69 x 10^-2 m^3.