Question

According to the ideal gas law, a 0.9054 mol sample of krypton gas in a 1.023 L container at 274.0 K should exert a pressure of 19.90 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Kr gas, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol. %

According to the ideal gas law, a 9.344 mol sample of oxygen gas in a 0.8267 L container at 500.1 K should exert a pressure of 463.8 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For O2 gas, a = 1.360 L2atm/mol2 and b = 3.183×10-2 L/mol. %

Compute the root-mean-square speed of N₂ molecules in a sample of nitrogen gas at a temperature of 164°C.

• m s^-1

Compute the root-mean-square speed of N₂ molecules in a sample of nitrogen gas at a temperature of 66°C.

• m s^-1

Answer 1

V = L.023 L Krypton n = 0.9054 moles T = 274.0K Pi = 19.90 atm (by ideal gas law) a = 2.318 L² atm/not? b = 3.978 x 10² L/mol. Pfan² (P + ar ) ( v- nb) = nRT = nRT p = - 2.18 katm (0.9054 0.9054 mét x 0.0821 K-atm x 274.4 mark 1.023 K - 0.9054 mye x 3.978x10 1998 Late xogor (1.0237 E IP = 18.82 atm % difference - = 19.40 -18.82 x 100 1990 5.427% Puw is 5.4271. less than Pideed nota Oxygen n = 9.344 moles. T- 500.1K a=11.360 6² atm v=0.8267 L P = 4 63.8 atm b7 3.183x10²4 P = 9-3 44 mot x 0. 0821 L-atm x 500 1K - 1.36 4² atm (9.344) and 2 0.8267 L - 9.344 mdx 3. 183x10 L (0.8267) 1² Pvw= $551.104 atm over 55).104-4638 x 100 = 18.82% Pro is 18.82% > Pideal 463.8

RMS speed 3 RT Nams - JM M= 28 g sud = 28 9xkg und JODD Molk a) M = 0.025 kg sud R = 8.314 I T=164°C = 164 + 273.15 = 437.15 k 3x8-314 h 43715 K me v 0.028 kg soll Vous ? v = 624.025 m/s 6) T=66°c = 339.15k. Von = 13x8.314 x 339.15k V 0.028 kg/nad ² 549, 65 m/s MS