Question

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 outside 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

Answer 1

Answer: The correct answer is Option B.

Explanation:

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.