A bicyclist finds the air pressure in his tires to be 230kPa at 12 degrees C. In the afternoon he checks again and the pressure is 241 kPa. Assuming volume does not change what is the afternoon temp?

asked by @lisan35 • over 1 year ago • PhysicsTemperature and Heat • 5 pts
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1 answer

In this problem, we have a case where an ideal gas is undergoing some sort of thermal process. Unless otherwise indicated, always assume a gas is ideal. The process we're told is ISOCHORIC (constant volume). Let's begin with the ideal gas law and figure out our conservation equation for an isochoric process (this is probably already listed in your book):

pV = nRT

V is a constant, so dividing p over, we get

V = nRT/p = consant


nRT_1/p_1 = nRT_2/p_2

Notice that since air isn't entering or leaving the tire that the number of moles, n, is a constant. R is just the ideal gas constant, so that doesn't change either. The only things to change in this problem are the pressures and the temperatures. Thus, we can cancel the nR from both sides of the above equation:

T_1/p_1 = T_2/p_2

What we want to find is the final temperature, T2. This means multiplying both sides by p2:

(p_2/p_1)*T_1 = T_2

Now, all we have to do is plug in our numbers. First, though, we need to make sure our values are in SI units. Pa is the SI unit for pressure, so that's fine, but K is the SI unit for temperature. Our initial temperature, in K, is

T_1 = 12 + 273 = 285 K


T_2 = (241 kPa / 230 kPa)*(285 K) = 298.6 K

Now, converting back into Celsius (which is likely how the question was meant to be answered, since T_1 was given in Celsius), we have

T_2 = 298.6 - 273 = 25.6 C

So, the final temperature is 25.6 degrees Celsius.

answered by @doug • over 1 year ago