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.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

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