The temperature of liters of a ideal gas was increased from 600 to 800 kelvins. If the volume remained constant and the final pressure was 30 atmospheres, what was the initial pressure?

Respuesta :

Answer:

the initial temperature is 22.5 atm

Step-by-step explanation:

The law describing pressure-temperature relationship is the Gay-Lussac's Law.

The Mathematical Form of which is:

=> [tex]\frac{P1}{T1} = \frac{P2}{T2}[/tex]

Where P1 = ? , P2 = 30 atm, T1 = 600 K and T2 = 800 K

=> [tex]\frac{P1}{600} = \frac{30}{800}[/tex]

=> P1 = 0.0375 * 600

=> P1 = 22.5 atm

So, the initial temperature is 22.5 atm