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. hw to do this sum?
Use the ideal gas relation
P1V1T2=P2V2T1
where temps are in Kelvins
and pressure is in absolute. Absolute pressure=guage+atmospheric.
To solve this problem, we can use the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the given values into the appropriate units. The volume is given in m³, and we can leave it as is. The gauge pressure is given in kPa, which is the same as 1000 Pa, so we convert it to Pa. The temperature is given in Celsius, so we need to convert it to Kelvin by adding 273.15.
Given:
Initial volume (V1) = 1.60 x 10^(-2) m³
Initial gauge pressure (P1) = 167 kPa = 167000 Pa
Initial temperature (T1) = 0.00°C = 273.15 K
Final volume (V2) = 1.69 x 10^(-2) m³
Final temperature (T2) = 27.0°C = 300.15 K
Now, let's use the ideal gas law equation to find the final gauge pressure (P2):
P1V1/T1 = P2V2/T2
Substituting the given values:
(167000 Pa) * (1.60 x 10^(-2) m³) / (273.15 K) = P2 * (1.69 x 10^(-2) m³) / (300.15 K)
Now, we solve this equation for P2:
P2 = (167000 Pa) * (1.60 x 10^(-2) m³) * (300.15 K) / ((1.69 x 10^(-2) m³) * (273.15 K))
P2 ≈ 186,429 Pa
Finally, to convert the gauge pressure from Pascals to kilopascals, we divide the result by 1000:
P2 ≈ 186,429 Pa / 1000 = 186.43 kPa
Therefore, 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) m³ is approximately 186.43 kPa.