Question

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is compressed to 30.0% of its original volume and the temperature is increased to 46.0°C. (a) What is the pressure in the tire (absolute)? Pa (b) After the car is driven at high speed, the tire's air temperature rises to 85.0°C and the tire's interior volume increases by 2.00%. What is the new pressure in the tire (absolute)? Pa

Answer 1

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

Given:

• $T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

• $T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

• $T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

• $V_1$ = The first volume of air inside the tire

• $V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

• $V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

• $P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

Assume:

• $P_2$ = The second pressure of air inside the tire

• $P_3$ = The third pressure of air inside the tire
• n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$.

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$.

Answer 2

Based on the general gas equation, the new absolute pressures in the tyre are;

• 380714.7 Pa
• 418882.1 Pa

What is the relationship between pressure, volume and temperature?

The relationship between pressure, volume and temperature is given by the general gas equation as follows:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

a) From the data provided:

P1 = 101,325 Pa

V1 = V

T1 = 10 °C = 283 K

V2 = 0.3V

T2 = 46 °C = 319 K

$P_2 = \frac{P_1V_1T_2}{V_2T_1}$

$P_2 = \frac{101325 \times 1 \times 319}{0.3 \times 283}$

$P_2 = 380714.7 \: Pa$

b) From the data provided:

P1 = 380714.7 Pa

V1 = V

T1 = 10 °C = 319 K

V2 = 1.02 V

T2 = 46 °C = 358 K

$P_2 = \frac{380714.7 \times 1 \times 358}{1.02 \times 319}$

$P_2 = 418882.1 \: Pa$

Therefore, the new absolute pressures in the tyre are 380714.7 Pa and 418882.1 Pa respectively.

Learn more about general gas equation at: https://brainly.com/question/20348074