Monday, April 16, 2012

If 1.00 of argon is placed in a 0.500- container at 29.0, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, and .


If 1.00 \rm mol of argon is placed in a 0.500-\rm L container at 29.0\, ^{\circ}C, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)?For argon,  a = 1.345 \;\rm (L^2\cdot atm)/mol^2 and b = 0.03219\; \rm L/mol.
Express your answer to two significant figures and include the appropriate units.
  P_{\rm ideal}-P_{\rm real}  =2.0  atm
Correct


P=\frac {nRT}{V-nb}-\frac {an^2}{V^2}

No comments:

Post a Comment