Someone me with the application problems:

1) P = 116.75 psi

2) V = 0.565 L.

3) V = 373.33 ml

Explanation:

1) Given, the pressure in tire is 105 psi at 25°C = 298 K

and then it is moved to another place where temperature is -5°C = 268 K.

Now, as the volume of tire remained constant, from gas law,

= constant.

⇒ P₁T₂ = P₂T₁ .

⇒ 105×298 = P ×268.      ⇒ P = 116.75 psi .

2) Given, balloon with volume 0.55 L at 22°C  = 295 K is placed outside where the temperature is 30°C = 303 K and also pressure is constant

⇒From gas laws, = constant

⇒ 0.55 × 303 = V × 295.           ⇒ V = 0.565 L.

3) Given, pressure in balloon changes from 0.80 atm to 1.2 atm

   and initial volume is 560 ml.

   As the temperature remained constant, from gas laws,

    PV = constant.

⇒ 0.80 × 560 = 1.2 × V.              ⇒ V = 373.33 ml.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )