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Chem 110 - Chapter 6 Assignment 1. Calculate the pressure of 1050 mmHg of argon in alm. lor o 10somntig 760mmtig = 798,000 attm 2. A 10.0L balloon contains helium gas at a pressure of 655 mmHg. Calculate the new pressure of helium, in mmHg and atm, when the balloon volume increases to 200 L assuming there is no changes in temperature V. - 10.0L V - 20.0L increases P. 655 mmHg P2? mootty Piv=PzV2 Pixe = P₂ 10.0 655 mmHg & zo. of - = P2 327.5 mm Hg = P2 328mmHg =P2 3. A gas has a volume of 4.00 L at 0.0°C. If the volume changes to 1200 ml., calculate the final temperature, in "C, to cause the volume of the gas to change to the following below assuming n and P does not change. T-273 K er 2 ? ? 1200mL OIL Vi= 4.00L V₂ = 1. 2L" Ti = 0 + 273= 27 1000 1.2 T2= Tive (2786) (1.24) 14.000) I'm # Iz r Vi = V2 x T Bla 82-173=1-191°C

Answer 1

0 Pressure of 1050 mm leg of argon in atm losoming & Honning 1050mning x latm = 1.38 atm 1. So, pressure of 1000 mming of argon in atm = 1.28 atm 327.5 According to Boyle's law P, = 655 mmily at constant tempersture v = 10L - PV = P2 V2 P = ? 655 mming X 10 L= P₂ X 202 l va = 20 P : * x mm ty Pressure in moly = P = 327.5 mm reg. Pressure in atm 327.5mmity x latin = 0.431 atm Homming According to charles law at constant hand p I = 4.002 T = 0.0°c= 273K a = 1200m2 = 1.202 | T₂ = ? 뷰 놀

4.00L 1.20 L 273K Tz T2 = 1.20 x 273 4.00 - 81.9k Tz - 8 (81.9 - 273)°C = -191.16 [ 12 = -191.16
question 1 is worng

Question 2 is right but pressure in atm missing.

Question 3 full right.