An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is (Gauge pressure is the difference between the actual pressure and atmospheric pressure.) After the car has been driven at high speeds, the tire temperature increases to (a) Assuming that the volume of the tire does not change and that air behaves as an ideal gas, find the gauge pressure of the air in the tire. (b) Calculate the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent.

This question is incomplete, the complete question is;

An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is 20°C (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

After the car has been driven at high speeds, the tire temperature increases to 55°C

(a) Assuming that the volume of the tire does not change and that air behaves as an ideal gas, find the gauge pressure of the air in the tire. (kPa)

(b) Calculate the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent. (kPa)

a) the gauge pressure of the air in the tire is 236 kPa

b) the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent is 205.33 kPa

Explanation:

Given the data in the question;

Gauge pressure at 20°C = 200 kPa

Absolute pressure at 20°C is; p = 200 + 101.3 = 301.3 kPa

Initial temperature T = 20°C = 20 + 273 = 293 k

Final temperature T" = 55°C = 55 + 273 = 328 K

Now, at constant volume, P"/P = T"/T  

P"T = PT"

P" = PT" / T

P" = PT" / T

we substitute

P" = ( 301.3 × 328 ) / 293

P" = 98826.4 / 293

P" = 337.29

so Gauge pressure at 55°C is;

⇒ P" - 101.3

⇒ 337.29 - 101.3 = 235.99 ≈ 236 kPa

Therefore, the gauge pressure of the air in the tire is 236 kPa

b)

Final Volume V'' = V + 10% of V  

V" = V + 0.1V

V" = 1.1V

we know that;

P"V"/T" = PV/T

P" = P (T"/T)(V/V")

we substitute

P" = 301.3 (328 /293 )(V/1.1×V)

P" =  × 1/1.1 × V/V

P" = 337.29 × 1/1.1  

P" = 306.628 kPa

so, absolute pressure at 55°C = 306.628 - 101.3 = 205.33 kPa

Therefore, the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent is 205.33 kPa

yup its correct its: c) b b

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer