Question

1.Gay-Lussac's Law

A gas container is initially at 47 mm Hg and 77 K (liquid nitrogen temperature.) What will the pressure be when the container warms up to room temperature of

25 ˚C?

2. A container of gas is initially at 0.500 atm and 25 ˚C. What will the pressure be at 125 ˚C?

3.A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20⁰ C, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15⁰ C?

4.A small research submarine with a volume of 1.2 x 10⁵ L has an internal pressure of 1.0 atm and an internal temperature of 15⁰ C. If the submarine descends to a depth where the pressure is 150 atm and the temperature is 3⁰ C, what will the volume of the gas inside be if the hull of the submarine breaks?

Answer 1

(1) Tofind the pressure Pressure is directly proporti onal with temperature P ccT Use the pressure -temperature law В Р. т, т, where P initial pressure Tinitial temperature P2 final pressure T2 final temperature P 47 mmHg T 77 K Р, 3 ? T 25 C+273 K = 298 K Ti T2 47 mmHg 77 K 298 K Cross multiplication are allowed 47 mmHg x 298 K P x77 K 47mmHg x 298K Р, - 77 K = 181.9mmHg Hence the pressure 181.9mmHg

(2) Tofind the pressure Pressure is directly proporti onal with temperature Р сст Use the pressure -temperature law Р Р. Т. Т, where P initial pressure Tinitial temperature P2-final pressure T2 final temperature P, 0.500 atm T 25 C+273 K = 298 K Р, 3 ? T, 125 C+273 K = 398 K T T2 0.500 atm Р, 298 K 398 K Cross multipli cati on are allowed 0.500 atm x 398K Px298 K 0.500 atm x398K Р, 298K 0.668atm Hence the pressure = 0.668atm

(5) To find the volume According to combinedgas law RVPV T2 where Pinitial pressure V initial volume Tinitial temperature - final pressure Р. V final volume T2 final temperature P 105 atm P2 0.65 atm V 50L Change the temaparature to kelvin T 200C273- 373K T2150+273= 123K T2 0.65 atm x V 105 atm x5,0L 373K 123K Cross multiplicati on 105 atm x 5.0L (123K) = (373K) 0.65 atm x V 105 atm x 5.0L (123K) (373K) 0.65 atm 105x5.0(123) (373) 0.65 2.66L Hence the volume 2.66L