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Elodia [21]
3 years ago
5

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.
Chemistry
1 answer:
Katarina [22]3 years ago
3 0

<u>Answer:</u> The new pressure will be 101.46 kPa.

<u>Explanation:</u>

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.

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