Question

the pressure in a 100ml container is 150.2 kpa at 500k. that pressure is adjusted to 101.3kpa, what is the new temperature​

Answer 1

The new temperature is 337.21 K when the pressure was reduced to 101.3 kPa from 150.2 kPa at 500 K.

Explanation:

Data given:

volume of the container = 100 ml

initial pressure on the container P1 = 150.2 kPa

Initial temperature of the container, T1 = 500 K

final temperature of the container, T2 = ?

Final pressure on the container P2 = 101.3 kPa

from the data provided, we will use Gay Lussac's law to calculate the final temperature:

$\frac{P1}{T1}=\frac{P2}{T2}$

rearranging the equation,

T2 = $\frac{P2T1}{P1}$

Putting the values in the equation:

T2 = $\frac{101.3 X 500}{150.2}$

T2 = 337.21 K

thus, the final temperature of the container is 337.21 K