hi, pls help me to solve this one using Gay-Lusasc’s Law

Question

1. The pressure of a container of helium at 25°C is

650 torr. If the sealed container is cooled down to

2°C, what will the pressure be?

2. A volume of the gas sample inside a container

exerts a pressure of 1.5 atm at 0°C. At what

temperature will the same volume of gas exert a

pressure of 4.5 atm?

Answer 1

Constant volume, pressure varies directly with the temperature for a given mass (moles) of gas.

This is Gay-Lussac's law.

P1/T1 = P2/T2

T1 = 25º + 273 = 298K

T2 = 2º + 273 = 275K

6.50 torr / 298K = P2 / 275K

P2 = 6.00 torr

For the second problem, you have...

P1 = 1.5 atm

T1 = 0 + 273 = 273K

P2 = 4.5 atm

T2 = ?

Use P1/T1 = P2/T2 and solve for T2