An automobile tire contains a certain volume of air at 30 psig and 70 °F. The barometric pressure is 29.50 Hg. If due to running conditions, the temperature of air in the tire rises to 160 °F. What will be the gauge pressure?

Assume that air is an ideal gas & that the tire does not stretch.

Make that psig for my last answer. (pounds per square inch, guage). The number is OK

Thank you very much

Well, it's always interesting to think about how hot air can really blow up your tires. Let's do some calculations and find out the gauge pressure!

First, convert the barometric pressure from Hg to psi. Since 1 Hg is roughly 0.491 psi, the barometric pressure is approximately 29.50 Hg * 0.491 psi/Hg = 14.48 psi.

Next, let's use the ideal gas law to find the final gauge pressure. The ideal gas law states that the pressure times the volume is equal to the number of moles times the gas constant times the temperature. Since the volume of air in the tire remains constant, we can write:

P1 * V = n * R * T1,

where P1 is the initial gauge pressure, V is the volume, n is the number of moles, R is the gas constant, and T1 is the initial temperature. We're looking for the final gauge pressure, so we'll rewrite the equation as:

P2 * V = n * R * T2,

where P2 is the final gauge pressure and T2 is the final temperature.

Now, let's get our equations together:

P1 * V = n * R * T1,
P2 * V = n * R * T2.

Since the number of moles, gas constant, and volume are the same for both equations, we can cancel them out:

P1 = P2 * (T1 / T2).

Now, let's plug in the numbers. P1 is 30 psi (gauge pressure at 70 °F), T1 is 70 °F, and T2 is 160 °F:

30 psi = P2 * (70 °F / 160 °F).

To find P2, we can rearrange the equation:

P2 = (30 psi) * (160 °F / 70 °F).

Calculating that, we get:

P2 ≈ 68.57 psi.

So, the final gauge pressure is approximately 68.57 psi. Keep in mind that this calculation assumes the tire does not stretch, but hey, it's always good to give your tires a little extra love now and then!

To determine the gauge pressure of the tire, we need to calculate the absolute pressure inside the tire at 160 °F and then subtract the barometric pressure.

To start, we convert the initial temperature of 70 °F to absolute temperature using the Rankine scale, which is the Fahrenheit equivalent of the Kelvin scale. The Rankine scale is obtained by adding 460 to the Fahrenheit temperature. So,
Initial temperature (Rankine) = 70 °F + 460 = 530 °R

Similarly, we convert the final temperature of 160 °F to absolute temperature:
Final temperature (Rankine) = 160 °F + 460 = 620 °R

Next, we convert the initial pressure of 30 psig to absolute pressure by adding the atmospheric pressure. The atmospheric pressure in psia (pounds per square inch absolute) can be obtained by converting the barometric pressure of 29.50 Hg to psia using the conversion factor. The conversion factor is 1 Hg = 0.491154 psia.
Atmospheric pressure (psia) = 29.50 Hg × 0.491154 psia/Hg ≈ 14.465 psia

Therefore, the initial pressure (psia) = 30 psig + 14.465 psia = 44.465 psia

Using the ideal gas law, we can relate the initial and final temperatures and pressures, assuming the number of moles and volume remain constant:
P1 / T1 = P2 / T2

Solving for P2 (the final absolute pressure),
P2 = (P1 * T2) / T1

Plugging in the values we already know:
P2 = (44.465 psia * 620 °R) / 530 °R ≈ 52.12 psia

Finally, to find the gauge pressure, we subtract the barometric pressure from the absolute pressure inside the tire at the elevated temperature:
Gauge pressure ≈ P2 - atmospheric pressure
Gauge pressure ≈ 52.12 psia - 14.465 psia ≈ 37.655 psig

Therefore, the gauge pressure of the tire at 160 °F will be approximately 37.655 psig.

At constant volume, the absolute pressure increaes in proportion to the absolute temperature increase.

P2/P1 = (344K)/(295K)= 1.166
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29.50 ihches Hg = 14.45 psia atmospheric pressure
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Original tire absolute pressure = 30.00 + 14.45 = 44.45 psia
Final tire absolute pressure = 44.45 * 1.166 = 51.83 psia
Final tire guage pressure = 51.83 - 14.45 = 38.4 psia