Question

Class-work ch.6 and ch. 10 Name 1) What is the final pressure (expressed in atm) of a 3.05 L system initially at 7wHg and 2 that is compressed to a final volume of 2.51 L at 273 K 1) 2) What is the pressure of a 1.0 L flask containing 0.60g of He at 25°C? (R-0.0821 L atm/mol K) 3) What is the volume of 28.0 g of nitrogen gas at STP? 3) 4) What is the final pressure of a system (atm) that has the volume increased from 0.75L to 11L with an initial pressure of 125 atm? 4) 0.85 1.SO .298 0.500 5) A rigid cylinder contains 2.00 liters of gas at a temperature of 25°C. If the pressure of this gas is changed from 0.500 atmospheres to 1.50 atmospheres, what will be the new temperature in Kelvin, reported to three significant figures) of the gas? (The volume is constant.) 5)_14.5K 447 25+ 273=298 1=0.500 2=1.50 298 x l.so o.soo

Answer 1

1) We know the relation between pressure, volume and temperature according to ideal gas equation as PV, PV₂ 2 -> 0 where P, & le are initial & final pressures respectively. v & are initial & final volumes respectively. T, & Tg are initial and final temperatures.

We know that 760 mm of Hg = 1 atm Given, P, = ( 124 ) atm Pa = ? V - 3.052 v 2.512 T, 296 Ta - 273k Substitute this values in eq. O, to get Pa value. (724) atm x 3.052 I = P2 x 2.516 2966 P2 = 1724) atm x 3.052 x 273K 2.51LX2966 273K 760

Pg - 1.067 atu P2 = 1.07 atm Hence, final pressure = 1.07 atm

2) use ideal gas equation, ie, PV = nRT 0 P = pressure = ? V volume = 1.0L n = no. of moles of gas (He) - Given mass of gas (He) molecular mass of life) 0.60g = 0.15 mol - 4 g/mol Ra 0.0821 L. atm. moi! K' T 250 = (25+293) K = 298 10 мо

in Substitute this values to get 'P' value. Px 1.0L = 0.15 mol x 0.0821 Latur mol.k x 298K - P. 0.15 md 20.0821 Latu. mol. KX 298K 1.0L P = 3.67 atm Hence, pressure is 3.67 atm

3 At STP pressure (pl= 1 atm Temperature - oc=273k Use ideal gas equation to get volume of nitrogen. A [PV = nRT 0 We know R - 0.0821 Latar molk' n = given mass of N₂ Molecular mass of N₂ 28.08 - Imol 28.05 mol substitule this values in , to get v. laton x V = 0.0821. L atm. mol. K ! x 293k 2) V- 0.0821 L.atm. mol. kx273K - 92.4L I am

4) Use Boyle's law equation, that is PV = P₂ V ₂ where P = initial pressure - 1.25 atm V = 0.75L (initial volume) y = final volume=1.12 P = ? (final pressure) 1.25 atm x 0.75L = 1.1 LX P2 - Pa = 1.25 atm x0.75L llL P = 0.85 atm

5) According to charle's law of constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature. P&T = P = KT constant = P = constant Given P. (initial pressure) = 0.500 atm Pq (final) = 1.50 atm T = 25C = (25+273 ) K = 298K To = ?

Substitute this values in 0, to get I value 0.500 atm – 1.50 atm 298K T2 =) T = 1.50 atm x 298K 0.500 atm T2 = 894k final temperature Ta = 894k = (894- 273)°C = 6216 To = 621c