Question

Charles' Law Worksheet Name Directions: Calculate the following using Charles' Law. Assume constant pressure, number of moles, and containers are able to expand/contract. If a 6.50 L balloon at 22.0 °C is put into a refrigerator at 4.0 °C, calculate new volume. A 500 L balloon is heated from 20.0-75.0 C, calculate the final volume. A sample of gas is cooled from 180 °C to 25 °C. The new volume is 15.5 L, what was the original volume? A sample of Neon is at 80.0 C with a volume of 63.5 L. If the volume was 45.0 L, what was the original temperature? Helium gas has a volume of 8.5 L at 225 C. If the sample was 5.00 L, what was starting temperature? I left a potato chip bag in my car in the summer. The volume is normally 250 mL at 19.0 C. I return to my car, what's the new volume at 60.0 °C. Some students believe that teachers are full of hot air. If I inhale 2.20L of gas at a temperature of 18.0 °C and it heats to a temperature of 38.0 C in my lungs, what is the new volume of the gas when I breathe it out?? Kr gas at 20.0 C was cooled from 63.5 "C. If the volume is 73.5 mL, what was the original volume? When you buy a balloon for a friend's birthday at a store, then go outside into the cold, why does the balloon deflate? What happens when you go back inside? Why don't you fill a balloon tightly while in the air conditioning if you are going to have it at an outside party in the summer? Why does the propane tank get cold while being used? Challenge: If you were to make a home-made thermometer out of gas in a tube with an expanding piston and you set the piston volume to be 2.00 L at 100 C, what would you wear outside if the volume is now 1.45 L? Shorts, long pants, or snow pants? 103

if someone could do out 5 of these questions, that would be great. thanks

Answer 1

As we you that Charles Law Vi - Vf I in mathematical form where viinitial volume Tia initial Temperature. Uf = final volume Tf a final Temperature Q :- If a 6.50 L balloon at 22.0°c is put into refrigerator at 4.0°C calculete new volume sole. Here, Vi =6. 5ol Ti z22°C 222 +273 = 295k Tf z 4°C 24+273 = 277K Therefore 6.50L z Vt. 295k 277k 277 K X 6.50L Uf = 2954 - 601033 L Vf=61L (Arso)

B. - A 0.500 L balloon ---.. . Sol Given here. Vi = o.sooL Ti = 20.0% z 2004 273 = 293.0k Vf = ? Tf = 75.0°C = 75.0+ 273 = 348.0k Therefore in ut ㅠ 파 o.sool 293.0k - Vf 3480k 20.5938 Uf = 348.0kx 0.500 L 293.ok Vf = 0.594 L A. :- A sample of gas is cooled from 180°C .... Solli- Given here Ti = 180°C = 180 +273 = 453 Tf z 25€ = 25+273 = 298k If z 15. 5L Viz?

Therefore, 15.5 298 Viz 15.5*453 223.56 L 298 Vi = 23.64 Am. SO Q.! - A sample of Neon ------ solution :- Given in the quest Tf = 80.0°C 2800+ 273 = 353.0k If = 63.5L Viz 45.0L TI ? Therefore, 1. Т 450L 63.5L Ti - 253.0k Ti = 353.0k X 45.0L 63.5L - Z250 kr Thus. Ti = 250-273 = -23°C Aus)

Qi- Helium gas has a volume - - Solo = 295.5k - Vfz 8.5L To = 22-5°C 222.5 +273 Vi=5.00L Ti= ? Thefer Therefore vi = vt 퓨 파 5.ool Ti 8.5L - 295.5k =) Ti= 5.00LX 2955k 8.5L = 173.8 K Thus. Ti= 173.8 - 273 z-99. 2°C (Aus.)