## Answers

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.)

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