A 1.50 liter flask at a temperature of 25°C contains a mixture of 0.158 moles of methane, 0.09 moles of ethane, and 0.044 moles of butane. What is the total pressure of the mixture inside the flask?

Question

anonymous · Answer

in ideal condition equal moles of each matter make same pressure and they don't differ hence we have 0.292 mole molecule 
 P=NRT/V=(0.292*8.31*298)/15*10^-4=482066 pa

anonymous · Answer

am i right?

anonymous · Answer

P = nRT / V = (0.158 + 0.09 + 0.044) mol x (0.08205746 L atm/K mol) x 
(25 + 273)K / (1.50 L) = 4.76 atm = 482 kPa = 3617 mmHg

australopithecus · Answer

@moradi312

Ideal gas law just describes a gas that does not interact with itself, it is not true to reality but it is a pretty good model for some gases

australopithecus · Answer

As it simplifies things

anonymous · Answer

Yes, and I propose that methane, ethane and butane are gases for which the ideal gas law is perfectly acceptable.  Their van der Waal's coefficients are relatively small (methane especially) and non-ideal behavior is more pronounced at high pressures, which is not part of this problem.

anonymous · Answer

Using van der Waal's equation (instead of the Ideal Gas Law) and the constants at http://www2.ucdsb.on.ca/tiss/stretton/database/van_der_waals_constants.html, I get a total pressure of 467.8 kPa = 4.62 atm = 3509, which is a 3.0% difference from the Ideal Gas Law results -- hardly a significant difference for schoolwork, especially since van der Waal's equation would have been specified in the question if that's what was expected.