es Asis often the case, the pressure exerted by the real gas sample is lower tha

Question

es Asis often the case, the pressure exerted by the real gas sample is lower than predicted by the ideal gas equation. Think About It Practice Problem ATTEMPT Using data from Table 11 1.5, calculate the pressure exerted by 11.9 moles of neon gas in a volume of 5.75 L at 25°C using (a) the ideal gas equation and (b) the van der Waals equation (Equation 11.14). Compare your results. Practice Problem BUILD Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50 Lat 32°C using (a) the ideal gas equation and (b) the van der Waals equation. Practice Problem CONCEPTUALIZE What properties of real gases prevent them from exhibiting ideal behavior? Explain why gases exhibit a greater degree of ideal behavior at very high temperatures andor at very low pressures.

Explanation / Answer

Given, n= 11.9 mol, V = 5.75 L, T = 250C = (25+273) K = 298 K

R = 0.08206 Latm /mol. K

a) Now, Using ideal gas equation we get,

PV= nRT

=> P = nRT/V

=> P = 11.9 mol x 0.08206 L atm/mol. K x 298 K / 5.75 L

=> P = (291.001172 / 5.75) atm

=> P = 50.6 atm

b) Now, using Van der Waals equation we get,

P + a(n/V)2 = nRT/V-nb

For Neon (Ne), a = 6.49 atm. L2/mol2 , b = 0.0562 L/mol

Now,

P =nRT/(V-nb) - a(n/V)2

=> P = 11.9 mol x0.08206 Latm/molK x 298 K / (5.75 L-11.9 mol x 0.0562 L/mol)-6.49 atm.L2/mol2(11.9 mol/5.75 L)2

=> P =( 291.001172 / 5.75-0.66878)atm - 6.49 atm.L2/mol2(141.61 mol2/33.0625 L2)

=> P= ( 291.001172 /5.08122) atm - 6.49 x (4.283) atm

=> P = 57.269 atm - 27.796 atm

=> P = 29.5 atm

Difference in pressure = 50.6 atm - 29.5 atm = 21.1 atm

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