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bogdanovich [222]
3 years ago
11

24. A sports ball is inflated to an internal pressure of 1.85 atm at room temperature (25 °C). If the ball is then played with o

utside where the temperature is 7.5 °C, what will be the new pressure of the ball? Assume the ball does not change in volume nor does any air leak from the ball A) 0.555 atm B) 1.74 atm C) 1.85 atm D) 1.97 atm
Chemistry
1 answer:
baherus [9]3 years ago
6 0

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}        (at constant volume)

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

Conversion factor:  T(K)=T(^oC)+273

P_1=1.85atm\\T_1=25^oC=(25+273)K=298K\\P_2=?atm\\T_2=7.5^oC=(7.5+273)K=280.5

Putting values in above equation, we get:

\frac{1.85atm}{298K}=\frac{P_2}{280.5K}\\\\P_2=1.74atm

Hence, the correct answer is Option B.

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