Question

A gas storage tank at a large industrial plant is designed to provide fuel under constant pressure. When the plant is shut down on Friday afternoon, the pressure in the tank is 115 kPa at 338 K. By early Monday morning the pressure in the tank had dropped to 108.75 kPa. What was the temperature on that Monday morning in K?

Answer 1

Answer: Temperature on monday morning is 319,5K

Explanation: This is an example of Gay Lussac law on gases which states that at constant volume, pressure and temperature are directly related, so if one raises the other one does as well.

The equation that follow the law is :

P1 / T1 = P2/T2 where 1 is the initial pressure and temperature and 2 is the final state.

In this equation, values have to be expressed in atm  for pressure and Kelvin for temperature.

$P1 = 115 kPa .\frac{1 atm}{101,3kPa} = 1,135 atm$

$P2 = 108,75 kPa . \frac{1atm}{101,3kPa} = 1,073 atm$

The missing value is T2 (monday temperature), so we have to clear it from the equation

T2 =P2.T1/P1

Replacing values:

T2 = 1,073 atm . 338K /1,135 atm

T2 = 319,5 K