Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

An automobile tire at [tex]\(30.0^{\circ} C\)[/tex] has a pressure of [tex]\(3.00 \text{ atm}\)[/tex]. The temperature decreases to [tex]\(-5.00^{\circ} C\)[/tex]. Assume that there is no volume change in the tire.

Formula to use: [tex]\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)[/tex]

Sagot :

Sure, let's solve this problem step-by-step.

### Given Information:
- Initial pressure of the tire, [tex]\( P_1 = 3.00 \text{ atm} \)[/tex]
- Initial temperature of the tire, [tex]\( T_1 = 30.0^{\circ}C \)[/tex]
- Final temperature of the tire, [tex]\( T_2 = -5.00^{\circ}C \)[/tex]
- We need to find the final pressure, [tex]\( P_2 \)[/tex], assuming no volume change in the tire.

### Formula to Use:
The problem states that we should use the formula for the relationship between pressure and temperature when volume is constant:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]

### Step-by-Step Solution:

1. Convert Temperatures from Celsius to Kelvin:

Temperatures need to be in Kelvin for the formula to work correctly, as the Kelvin scale is the absolute temperature scale used in gas law equations.
[tex]\[ T_1(K) = T_1(^{\circ}C) + 273.15 = 30.0 + 273.15 = 303.15 \text{ K} \][/tex]
[tex]\[ T_2(K) = T_2(^{\circ}C) + 273.15 = -5.00 + 273.15 = 268.15 \text{ K} \][/tex]

2. Use the Formula to Find the Final Pressure:

Rearrange the equation [tex]\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)[/tex] to solve for [tex]\(P_2\)[/tex]:
[tex]\[ P_2 = P_1 \times \frac{T_2}{T_1} \][/tex]

3. Substitute the Known Values:

[tex]\[ P_2 = 3.00 \text{ atm} \times \frac{268.15 \text{ K}}{303.15 \text{ K}} \][/tex]

4. Calculate the Final Pressure:

[tex]\[ P_2 \approx 3.00 \times 0.884 = 2.65 \text{ atm} \][/tex]

So, the final pressure in the tire, when the temperature decreases to [tex]\(-5.00^{\circ}C\)[/tex], is approximately [tex]\(2.65\)[/tex] atm. The intermediate values used, specifically the converted temperatures, are [tex]\(303.15\)[/tex] K and [tex]\(268.15\)[/tex] K.