A bicycle tire is filled with air to a pressure of 100 psi at a temperature of 19C. Riding the bike on asphalt on a hot day increases the temperature of the tire to 58C. The volume increases by 4%. What is the new pressure of the tire?
Convert the temps to Kelvins, then use the combined gas law.
About 108?
i am having problems with my physical science problem : combined gas law
Here goes: A gas has a pressure of 340 kPa at avolume of 3.20L. What happens to the pressure when the volume is increased to 5.44L? The temperature does not change. Please help. Thank you....
at a pressure of 780mm Hg and 24.2 c,a certian gas has a volume of 350.0 mL .what will the volume of this gas under STP.
To solve this problem using the combined gas law, we can use the equation:
(P1V1)/T1 = (P2V2)/T2
Given:
P1 = 340 kPa
V1 = 3.20 L
V2 = 5.44 L
T1 = T2 (temperature does not change)
We can rearrange the equation to solve for P2:
P2 = (P1V1T2)/(V2T1)
Substituting the given values:
P2 = (340 kPa * 3.20 L * T1) / (5.44 L * T1)
The T1 cancels out:
P2 = (340 kPa * 3.20 L) / 5.44 L
Simplifying the equation:
P2 = 200 kPa
Therefore, the new pressure when the volume is increased to 5.44L is 200 kPa.
To solve this problem, you can use the combined gas law:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
In this case, the temperature does not change, so T1 = T2.
Given:
P1 = 340 kPa
V1 = 3.20 L
V2 = 5.44 L
T1 = T2 (no change in temperature)
Now we can plug in the values into the combined gas law equation:
(340 kPa * 3.20 L) / T = (P2 * 5.44 L) / T
Since T1 = T2, we can simplify the equation further:
(340 kPa * 3.20 L) = (P2 * 5.44 L)
Now rearrange the equation to solve for P2:
P2 = (340 kPa * 3.20 L) / 5.44 L
P2 = 200 kPa
Therefore, the final pressure, when the volume is increased to 5.44 L with no change in temperature, would be 200 kPa.