Question

Ideal Gas Law PV = nRT Gas Laws Formulas Combined Gas Law (P1)(V1) Ti (P2)(V2) Tz = Dalton's Law of Partial Pressures P = P, +P+P, + ...etc. K="C + 273 3. A 5.0 L sample of gas has a pressure of 1200 torr. What is the new pressure if the temperature is changed from 220 K to 440 K and the volume increased to 20.0 L ? 4. A 10.0 mL sample of gas has a pressure of 1.20 atm and a temperature of 150 °C. What is the volume when the pressure is 0.90 atm and the temperature is -40.0°C? af Nao 5. What is the volume of 0.25 mole of nitrogen gas at 0°C and 1.5 atm? 6. What is the pressure of 1.0 mol of a gas at 5.0 L and 18 °C ? 7. A mixture of nitrogen and oxygen has a total pressure of 840 mmHg. If the oxygen has a partial pressure of 510 mmHg, what is the partial pressure of nitrogen in units of atm ?

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Answer 1

Initied volume (V)= s.ol, final volume (2) = 20.0L Ini Hal temperature (7) - 220k, final temp. (T) = 440k Initial pressure (P.) = 1200 derr. using PV = Pete Pa = P. 1 1 2 yok 1200 horr & S.olx 220 K X 20.0L = 6oo torr. New Pressure is boo tor. Initial volume (V.) = loom, final volume = ? In hieel temperature (7,7 = 1501 = 423 k final temperature (Tr) = You= 233K Initial pressure (P.) = 1.2 atm final Pressure (P) = 0.9ahm T = 12-12 V₂ = PV, Tz V 7.34 mL 1.2mm x 100ml x 233K 423k * 0.9 ahm Hence, final volume = 7.54mt at T = -40C & P = 0.gahm, (6 = 0.25 mole, T=0°C = 273K, P=1.5 atm PV = nRT = = ART V= 0.25 mol x 0.082 Lahm k mol x 873K 1.5 am V= 3.73 L Hence 3.73L is the volume of 0.25 mod Ny at o'c & 1.5 am

n=1.0 mod, v= S.OLT = 1806 = 2915 P= nRT = lomol x 0.082 Lahm Kimool x 2915 - S.OL = 4.77 atm. Honce, 4.77 atm is the and 180C. pressure of 1.0 mod of gas at Sool Total pressure (PT) = 840 mmHg Partial Pressure of Oz (Poe) = slommig. Pasitial Pressure of N₂ (PN) = Px - Por = 840mmile - Slommitte = 330mtlg. 330 atm lam= 760 mmity] = 0.434 am Hence, Partial Pressure of N₂ is 0.434 alm. 760