Question

A gas has a pressure of 810 kPa at 227°C. What will its pressure be at 27°C?

Answer 1

486 kPa will be the pressure of gas at 27 degrees.

Explanation:

Data given :

Initial pressure of the gas P1 = 810 kPa

initial temperature of the gas T1 = 227 degrees OR 500.15 K

final pressure of the gas =?

final temperature of the gas = 27 degrees OR 300.15 K

Gay Lussac's law is used to calculate the pressure of the gas at 27 degrees or 300.15 K

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

P2 = $\frac{P1T2}{T1}$

 Putting the values in the equation:

P2 = $\frac{810 X 300.15}{500.15}$

P2 = 486 KPa

486 Kpa is the pressure of the gas when temperature was reduced to 27 degrees.