Question

1. A gas has a volume of 1.57 L at 2.95 atm. What pressure is required to reduce the volume to 1.25 L at the same temperature? at ga 2. A gas has a volume of 3.75 liters at 12°C, what is the volume at 25 C if the pressure remains constant? wil 3. What is the pressure of a gas in a 375 mL container, if it has a pressure of 219 Torr in a 525 mL container at the same temperature? 4. What is the pressure of a gas in a 2.00L container at 35 C if it has a pressure of 117 mm Hg in a 1.75L container at 45°C?

Answer 1

1. Initial Pressure, P, = 2.95 atm Initial volume, v = 1.57 L Final volume, 2 = 1.25 L Final Pressure, P = 2 to be calcu. lated 7 Now as the temperature is const- ant, we can use Boyle's Law as below - P, V = P2 V2 a P2 = Piv - 2.95 atm x 1.57 1.25 X = 3.70 atm [three signific - aut figures] So, the processure required = 3.70 atm Aus 2. Initial volume, u = 3.75 L Initial Temperature T, = (12+273)k = 285k

I final Temperature, 12 = (25 +273)* = 298K Final volume, v = 2 to be carcular ted ] As the So, from pressure remain constant, Charl's law - 3.75L x 298K 285 = 3.92 L three significant figures] Thus the volume become = 13.92 L Ans 3. Given, P = 219 torn V = 525 mL

2 = 375 ml P2 = ? to be calculated] Here also the temperature is fixed. Thus, froom Boyel's law - P, V = P2 V2 a P₂ = Piri - 219 torr x 525 ml 375 ml = 306.6 torn So, the process wre of the gas would be = 1306.6 for Ane - A. Given, I P, = 117 mm Hg V,=1.75 L T = (45 + 272) K = 318k - Again -

P2 = ? [to be calculated] v2 = 2.00 L ; T2 = (273 +35)k = 308 K Froom combination law of Boyle's and charl's - vi - Pq Vz Ti Ta a P = P VT 2 - T X v2 117 mm Hg x 1.75 X x 308K 3184 x 2.000 = 99. 1 mm Hg Thus, the pressure would be = 199 1 mm Hg) Are
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