Question

A cylinder has a volume of 10.5 L and the pressure of the gas inside is 0.92 atm. By moving the piston, the volume increased to 15.4 L. What is the new pressure? O 0.84 atm 1.3 atm 1.4 atm O 0.63 at

Answer 1

Solution 1: Given Data Intal volume of cylinder = 10.5L Intial presure of gas = 0.92a1m Final volume of gas in cylinder = 15.4L final pressure of gas in cylinder = ? - By applying Boyle's Law O pat PV = Constant Eben PV, = constant - BzV2 = Constant E no, Piv, = BV Prave are the intial pressure & volume of gas respectively Papavą are final pressure A volume of gas respectively, 20.5LX 0.92 atm = Pex 15.42 - Pe: 10.54 x0.9291m 15.4L E ps = 0.63 atm = Seg oftion Tuth is correct. Need pressure is o.f3 atm. - Solution : 2: Given Data Irrtial volume of ballon = 15.5 L Intial Temperature of hallon : 35°C or 9 43+35°= 308 K final temperature of ballo: 5.2°C or 273+52:279.2K CS Scanned imith volume of ballon = 2 CamScanner

by applying Charlets Law V constant Intial Vit Ti are the volume of T h e temperature of ballon respectively 1. V Tf are final volume & temp -erature of ballon respectively. 15.5L - S V 508K 278.2K 15.5L X 278.2K 7 308k V = 14:02 So, option and correct & New volume Solution - 3:- Given Data 14.02. Volume of gas = 11.22 Temperature of gas: 14°C or 273+14°C = 2875 pressure of gas = 1.13 atm R= 0.0820574 Latm molt -1 No of moles of gas n = ? By applying ideal gas equation PV = nRT Scanned with CamScanner

- So, PV RT 1.13 atm x 11.2L 57.0820574 [adm mol ?k** x 987 % na 0.537 moles. So, option 10th is correct. Solution 4i- Given Data Total pressure (0)= 740 tony = 0.944'atm No of moles of NQ = 1:3 mol No.of moles 4 0g = 0.33 mo) NO of moles of AY 0:0 Gimel So, Mole fraction = Moles of individuel species Total moles present in mixture Xi- - ni Eni XNg = a 1.3 1.3 13+ 0.33+0'OGI XNg = 0.768 Similarly = 0: 195 0.33 131033 to 061 006 Scanned with CamScanner 137033 +061

By applying relallim of meled fraction and partial pressure of you Pie Yi Platale della Es, where Piz Partial pressure of individueel gos in misture fi a role fraction of individlasul year in mixture Proteal - Total pressure. son Pro : 0.468 x 440 torr - 0768X 0.918 aten - 0.75 atm Simikrly, Po = 0.195 x 0.978 atm = 0.19 atm - 0.036 X0:978 atm. = 0.035 atm So, option and is correct and partial pressure of HA Og & Ar are 0.75, 0.19 & 0.035 admi respectively. Scanned with CamScanner