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An automobile tire is filled with air at a pressure of 27.0 lb/in2 at 25°C. A cold front moves through and the temperature drops to 5°C. Assuming no change in volume, what is the new tire pressure?

Respuesta :

Answer:

[tex]P_2=25.18\ lb/in^2[/tex]

Explanation:

Given that

P₁=27 lb/in² ,T₁=25⁰C=273+25 = 298 K

T₂=5⁰C = 273 + 5  = 278 K

Lets take final pressure = P₂

We know that if volume of the gas is constant then we can say that

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

[tex]P_2=P_1\times \dfrac{T_2}{T_1}[/tex]

Now by putting the value in the above equation

[tex]P_2=27\times \dfrac{278}{298}\ lb/in^2[/tex]

[tex]P_2=25.18\ lb/in^2[/tex]

Therefore the new tire pressure will be

[tex]P_2=25.18\ lb/in^2[/tex]

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