Question

The following gases are stored in identical containers at the conditions specified. Which of the gases has the HIGHEST pressure?

15 g He at 250K

96 g O2 at 200K

71 g Cl2 at 150K

66 g CO2 at 300K

Answer 1

Here we should use the ideal gas law equation that iss. PV = nRT where, P= Preasure of the gas v=volume of the gove n = Number of mole of the gars R = Molari gara conotant T= temperature of the gars Again, nam where, m=mars of the garo M = molar mano of the goro

Harre in this problem given, all garen are stored in identical containers Therefore, volume of all gamen are name. het volume of the container =VL For, 159 of He at 250K T= 250K m = 15g R = 0.0821 L-atm. motta! V=VL And molari mann (M) of He = 4:00 2602 g/mol Now, pV - MRT >> P = MRT + P = (159) x (0.0821 L-atm, mol'k') x (250m) (4.00 26 02 g/mol) x (VL) - 317. 875 4.00 2602 xv atm. M MV P= 76.92 atm

For 96g of other rook recoure of He gas in 76.92 atm T= 200k m= 96g R = 0.0821 L-atmimoila! v=VL Molan mars (M) of 9 = (2x 15.999) g 1 mot = 31.998 g/mol Now, PV=MRT M P= m RT м*, pa (0.0821L-atm, móil. a')* (2006) 31.998 g mot 1 (NL) 9 og 49.25 atm =) po 1576-32 xk atm 31.998 P= 49.26 atm Therefore, Pressure of a garo is 49 For 71g of cla at 150k T = 150k m = 71g R = 0.0821 L-atm. moł' V - VL Molari mana (M) of clo = 2(35.4 5.3 g/mol) ? = 70.906 g/mol Now, PVE MRT p= m RT + P- 719 (0.0821 L-atm. możl. ') x (1505). 70.906 gimot T (VL) - 874.365 x katm P = 70.906 T PF 12.33 atm. Therefore, pressure of cha gan in 12 33 atm.

m- For, 669 co2 at 3ook T= 300k R = 0.0821 L-atmimot' ' v=vL Molar mann (M) of co= 44.01g/mol Now, PV=MRT = pa m x RT P= < (0.0821 L-atmi mot' a' ) x (300K) 7 - 44.01g Imola (VL) = P = 1625:58 x 's atm 44.01 669 44 mal p= 36.94 atm Therefore, prenoure of coganin 36.94 atm. An volume in name for all the gaver, so it in He gon had the highent pressure. clear that