Question

Air was put into an automobile tire when the tire was cold, at 5.00 C. The tire's air pressure was 3.04 atm. Later, the weather warms up. Also, the automobile is driven, further warming the tire. What will the tire's pressure be when the temperature is 25.0 C?

Answer 1

Answer:

P2=3,26 atm

Explanation:

Considering that this gas behaves like an ideal gas, the equation used in noble gases is:

PxV = nxrxT

p is pressure, v is volume, n is number of moles, r is a constant of noble gases whose value is 0.082x10exp23 as long as you use it in the equations it has the same value (that is why it is called constant) and t would be the temperature.

Because v, r and n are values that did not change throughout the reaction of the gas, we are only going to take into account the p and t that did change throughout the reaction and that is why there is an initial and final pressure and a final and initial temperature.

So:

P1 / T1 = P2 / T2 (where value 1 refers to start, and 2 to end)

Regarding this equation, since we want to know the value of P2, which is the FINAL pressure of the gas, that is, what pressure did it achieve with the reaction, the equation that we would use would be solving for P2:

P2 = (T2 / T1) xP1 = 3.26 atm

Answer 2

Answer:

$P_{2} = 3.259\,atm$

Explanation:

Let suppose that air inside the tire behaves ideally. The equation of state for ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

As tire can be modelled as a closed and rigid container, there are no changes in volume and number of moles. Hence, the following relationship is constructed:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{298.15\,K}{278.15\,K}\cdot (3.04\,atm)$

$P_{2} = 3.259\,atm$