Question

the pressure in a sealed plastic container is 108 kPa at 41 degrees Celsius. What is the pressure when the temperature drops to 22 degrees Celsius? Assume that the volume has not changed.

Answer 1

Answer: The new pressure will be 101.46 kPa.

Explanation:

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

The equation given by this law is:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are initial pressure and temperature.

$P_2\text{ and }T_2$ are final pressure and temperature.

We are given:

By using conversion factor:   $T(K)=T(^oC)+273$

$P_1=108kPa\\T_1=41^oC=314K\\P_2=?kPa\\T_2=22^oC=295K$

Putting values in above equation, we get:

$\frac{108kPa}{314K}=\frac{P_2}{295K}\\\\P_2=101.46kPa$

Hence, the new pressure will be 101.46 kPa.