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On a cool morning, when the temperature is 11 ∘C, you measure the pressure in your car tires to be 30 psi . After driving 20 mi on the freeway, the temperature of your tires is 47 ∘C. Part A What pressure will your tire gauge now show?

Respuesta :

Answer:

The answer is 33.8 psi.

Explanation:

We will use the ideal gas law for the solution.

The gas law equation is:

[tex]PV=mRT[/tex]

As the volume of the gas V is constant along with mass m and ideal gas constant R

Therefore rearranging the equation, it becomes:

[tex]V/mR=T/P = constant[/tex]

which means [tex]T1/P1 = T2/P2[/tex]

Now we will put the values from our question.

[tex]T1 = 11 C = 284.15 K\\P1 = 30 psi\\T2=47 C=320.15\\P2 = required[/tex]

Solving the equation for P2

[tex]P2 = P1*T2/T1\\P2 = 30*320.15/284.15\\P2=33.8 psi[/tex]

Cricetus

The pressure will be "50.36 psi".

We know,

  • Atmospheric pressure = 14.7 psi

then,

  • Initial absolute pressure, [tex]P_1 = 30+14.7[/tex]

                                                        [tex]= 44.7[/tex]

  • Initial gauge temperature, [tex]T_{gi} = 11^{\circ}C[/tex]
  • Initial absolute temperature, [tex]T_1 = 11+273.15[/tex]

                                                               [tex]= 284.15 \ K[/tex]

  • Final gauge temperature, [tex]T_{gf} = 47^{\circ} C[/tex]
  • Final absolute temperature, [tex]T_2 = 47+273.15[/tex]

                                                              [tex]= 320.15 \ K[/tex]

We know the relation,

→ [tex]\frac{P_2}{P_1} = \frac{T_2}{T_1}[/tex]

or,

→ [tex]P_2 = \frac{P_1 T_2}{T_1}[/tex]

By substituting the values, we get

        [tex]= \frac{44.7\times 320.15}{284.15}[/tex]

        [tex]= 50.36 \ psi[/tex]

Thus the above answer is right.

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