Question

You have a 3.00-liter container filled with N2 at 25°C and 1.75 atm pressure connected to a 2.00-liter container filled with Ar at 25°C and 2.85 atm pressure. A stopcock connecting the containers is opened and the gases are allowed to equilibrate between the two containers. What is the density of the final gas mixture? Assume ideal behavior.

Answer 1

Given, container of An container volumeli 2.00 L Temp (t) = 25° c = 290 Pressure - 2.85 atm Volume = 3.00 liter Temperature = 25°c = 298% Pressure - 1.75 atm. According to Ideal gas equation, PV = nRT N2 where no no. of moles of a gas . 1.75 x 3 = MN x RXT = M = 5.25 similarly, PY = NAD XR XT nar= no. of moles of argon . 2x 2.85 = NAY X RXT Mar = 5.7 RT RT when the stop cock equillirium, and is opens the hence, the gases no. of come to moles of gas, n = NAY & MN2

mar = 5.7 0. 0821 x 298 = 0.232 .: May = 5.25 0.0821x 298 = 0.214 mans of no. Na present = 0.2144 28 = 6g present man of As - 0.232x 40 = 9.319 Mol. ct of the mixture nodar volume (6) - density (d) 69 + 9.31g) => 7 c d (3.00 +2.00) d = 3.06 3.06 g/L