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sesenic [268]
3 years ago
14

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?
Physics
1 answer:
Softa [21]3 years ago
7 0

Answer:

P_2=25.18\ lb/in^2

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

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=P_1\times \dfrac{T_2}{T_1}

Now by putting the value in the above equation

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

P_2=25.18\ lb/in^2

Therefore the new tire pressure will be

P_2=25.18\ lb/in^2

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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.

Complete step-by-step answer:

In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.

Square of 12 = 122=144 and square of 12 = 132=169

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