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# PHYSICAL CHEMISTRY I CH 353

UT

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This 4 page Class Notes was uploaded by Brady Spinka on Monday September 7, 2015. The Class Notes belongs to CH 353 at University of Texas at Austin taught by Xiao Zhu in Fall. Since its upload, it has received 54 views. For similar materials see /class/181875/ch-353-university-of-texas-at-austin in Chemistry at University of Texas at Austin.

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Date Created: 09/07/15

HW 1 Key 1212011 6 Q Research in surface science is carried out using ultrahigh vacuum chambers that can sustain pressures as low as 103912 torr How many molecules are there in a 100cm3 volume inside such an apparatus at 298K What is the corresponding molar volume l7 at this pressure and temperature A Assume Ideal gas behavior PV zn Inserting the appropriate constants for the units 10 12t0rr100cm3 82058 cm3 atm mol 1 K 1760 torr atm 1298 K n Gives 324x104 molecules in 1 cubic centimeter inside the apparatus The molar volume 17 2lt where n is the number of moles or our solution from above divided by Avogadro s number is V 13900 mg 1 86 1019 3 r1 I a a 538 10720 mol m quotm 11 Q It takes 03625 g of nitrogen to fill a glass container at 2982K and 00100 bar pressure It takes 09175 g of an unknown homonuclear diatomic gas to fill the same bulb under the same condidtion what is the gas A Two things to realize First at the same T P and V the number of moles of gas is the same Given its mass and a way to calculate the number of moles we can get the molar mass of the homonuclear diatomic gas 03625 g 72 nN2 2801359m0171 1294 10 mol The molar mass of the unknown compound must be 09175 g M 1294 102 mol 70903 g mar1 Dividing that by two gives us the molar mass of one atom of our gas which is that of chlorine so the gas is Clz 13 Q Use the van der Waals equation to plot the compressibility factor 2 against P for methane for T180K 189K 190K 200K and 250K Hint Calculate 2 as a function of l7 and P as a function of l7 and then plot 2 versus P A For methane a23026 dm sbarmol392 and b0043067 dm3mol39l By definition Z P17 RT and the Van der Waals equation of state is P RT 1 i7 b V2 Substitute P into top equation and create parametric plot of 2l7 versus Pl7 for suggested temperatures gives a plot like the one below at High T molecular attraction is less important Z 300 Pressure bar 25 Q Another way to obtain expressions for the van der Waals constants in terms of critical paramters is to set 6P6l7T and 62P6l72T equal to zero at the critical point Why are there quantities equal to zero at the critical point Show that this procedure leads to Equation 1612 and 1613 A These Values are equal to zero at the critical point because the critical point is an inflection point in a plot of P versus V at constant Temperature RT P a V b 172 2a azPavz ZRT 6a T b3 V4 If both zero at critical temp RTCl73C moi b2 ZRTCWC moi 13 Multiply first derivative by 2V gives ZRTCWC mini 12 Setting these two equal to one another gives 4ai717 b2 6aUZ bf 3b 12 Substituting this into RTCl73 2al7c b2 from above gives RTC3b3 2a3b b2 T 8th2 8a 27b3R 27bR 8L1 Substituting 3b 17 and TC NH into van der Waals equation to find Pc RTE a 8aR a a C Z b 762 27bR3b b 3b2 2712 Using these expressions for the constants Equation 1612 follows 59 Q The isotherm compressibility l is defined as 1 K E 76V6PT Show that For an ideal gas A For an ideal gas P RTl7 taking the partial derivative of both sides with respect to P gives RT 172 6V6PT andso 1 av ap l7 1 KV T RT P The answer for read math chapter H can be found in math chapter H

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