An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is compressed to 28.0% of its original volume and the temperature is increased to 36.0°C.
a) What is the tire pressure?
b) After the car is driven at high speed, the tire air temperature rises to 85.0°C and the interior volume of the tire increases by 2.00%. What is the new tire pressure (absolute) in pascals?
Solution to above questions
I need a solution to the above questions
To find the tire pressure in both situations, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
However, we need to make sure that all the values are in the correct units. The ideal gas constant is typically given in units of J/(mol·K), but since we want the answer in pascals, we'll use the value R = 8.314 J/(mol·K).
a) What is the tire pressure after compression and temperature increase?
First, let's calculate the initial volume of the tire. We know that the air is compressed to 28.0% of its original volume, so the new volume (V') is given by:
V' = 0.28 * V
Next, we need to convert the temperatures to kelvin by adding 273.15:
T_initial = 10.0°C + 273.15 = 283.15 K
T_final = 36.0°C + 273.15 = 309.15 K
Now, let's plug in the values into the ideal gas law equation. Since the number of moles (n) and volume (V) are constant, we can write:
P_initial * V = P_final * V'
Simplifying, we get:
P_final = (P_initial * V) / V'
Substituting the known values, we have:
P_final = (P_initial * V) / (0.28 * V)
Since the volume cancels out, we are left with:
P_final = P_initial / 0.28
Therefore, the tire pressure after compression and temperature increase is 3.57 times the initial pressure.
b) What is the new tire pressure after temperature increase and volume change?
Again, let's convert the temperatures to kelvin:
T_final = 85.0°C + 273.15 = 358.15 K
To find the new volume, we need to account for the 2.00% increase. We can calculate the new volume (V'') by multiplying the initial volume (V) by 1 + (2.00/100):
V'' = V * (1 + 2.00/100)
Now, let's use the ideal gas law equation again:
P_initial * V = P_final * V''
Simplifying, we get:
P_final = (P_initial * V) / V''
Substituting the known values, we have:
P_final = (P_initial * V) / (V * (1 + 2.00/100))
Simplifying further, we have:
P_final = P_initial / (1 + 2.00/100)
Therefore, the new tire pressure after the temperature increase and volume change is approximately 0.98 times the initial pressure.
P1V1/T1=P2V2/T2
P2= P1V1T2/T1V2= you do it
V2=.28V1