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vesna_86 [32]
3 years ago
6

An automobile tire having a temperature of −1.5 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 26 lb/in2 . What

would be the gauge pressure in the tire when its temperature rises to 40◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and that the atmospheric pressure remains constant at 14.7 lb/in2 . Answer in units of lb/in2 .
Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:32.22\ lb/in^2

Explanation:

Given

Temperature on the hot day T_i=-1.5^{\circ} C\approx 271.5\ K

Gauge Pressure P_i=26\ lb/in^2

When Temperature rises to T_f=40^{\circ}C\approx 313\ K on a hot day then

absolute pressure  is let say P_f

Using ideal gas equation we can write

\Rightarrow \frac{PV}{T}=constant

as volume is constant therefore

\Rightarrow \frac{P}{T}=constant

\Rightarrow \frac{26+14.7}{271.5}=\frac{P_f}{313}

\Rightarrow P_f=46.92\ lb/in^2

Gauge Pressure is given by

\Rightarrow 46.92-14.7=32.22\ lb/in^2

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The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

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PV = nRT...............................  Equation 1

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n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

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D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

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Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

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13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

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28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

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