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4. Consider 1.0 mol of ethane in a 22.414 dm3 vessel at 273.15 K and calculate the pressure when the gas behaves as a a. Perfect gas. b. van der Waals gas.

          4. Consider 1.0 mol of ethane in a 22.414 dm3 vessel at 273.15 K and calculate the pressure when the gas behaves as a
a. Perfect gas.
b. van der Waals gas.
        

Added by Kimberly P.

Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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4. Consider 1.0 mol of ethane in a 22.414 dm3 vessel at 273.15 K and calculate the pressure when the gas behaves as a a. Perfect gas. b. van der Waals gas.
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Transcript

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00:01 This particular problem asks us to calculate the pressure of a gas, particularly ethane, according to the ideal gas law and the vanderbalds equation.
00:15 The vanderval's constants are given in a table.
00:20 So we're using two equations here to calculate pressure.
00:24 The ideal gas law is simply pv equals n over t, nrt, pardon.
00:34 But because we have one mole, we're simply.
00:39 Going to write rt and this becomes a molar volume with the line above.
00:45 That simply means the units instead of liters would be liters per mole.
00:52 So to solve for pressure, we are going to plug in rt divided by v molar.
01:03 I'm going to use an r that involves leaders and bars because in part b, we will also use the same units for comparison.
01:12 And i will show you why.
01:16 So i'm going to use r 0 .08 3145 liter bar per mole k.
01:37 I'm going to multiply this by 373 kelvin and divide it all by our molar volume.
01:47 Now notice that we have liters and bars and our r constant here.
01:53 So that means we need liters for our volume.
01:56 Our volume is 200 mils.
02:00 If we divide that by a 1 ,000.
02:02 Or simply think of decimal point moving over by 3 that is 0 .2 00 liters so we can plug in that value 0 .2 00 liters and when we solve this we can recognize here that we in fact only have one significant figure so our answer is not exactly 200 but the reality is 200 milliliters just only has one significant figure.
02:44 We're assuming that these trailing zeros are not significant.
02:47 So really, this is 0 .2 liters, and that's all we know.
02:54 We don't have enough information for the rest of these.
02:59 So the answer, rounded to 1 sig figs, is 200 bar.
03:07 So now to answer the second part.
03:13 The second part, we actually need to consider what gas it is.
03:20 The ideal gas law in part a, this applies to all gases.
03:26 So whether or not this was ethane really didn't change this equation.
03:32 But the vanderval's equation does.
03:36 The vanderval's equation takes into account what the real gas is.
03:44 And each gas will deviate from ideality in its own specific way.
03:50 To solve the vandervald's equation, we need to know the values for two co -oferencing.
03:55 Coefficients a and b.
03:59 These are the vandaval coefficients.
04:02 We can find these from any table that includes ethane.
04:13 From the table i've consulted, i know that my values for a.
04:22 This value is 5 .562 liters squared bar per mole squared.
04:35 And b is 0 .06380 liters per mole.
04:57 So what do these numbers actually mean? so this is the table specifically for f -f -a, c2h6.
05:11 Each of these numbers gives us a correction factor.
05:15 B, if we just look at the units, we can see that b is a correction for volume.
05:20 This is the actual volume taken up.
05:25 Or occupied by the molecules.
05:32 In the ideal gas law, we are assuming that the molecules do not take up any space, but in fact they do.
05:40 This correction is for internal molecular forces and attractions between the gaseous molecules.
05:54 The combination of those will tell us how far we differ from ideality of 200 bars.
06:02 So the vanderval's equation looks very similar to the idealist.
06:07 Ideal gas law, but we have to correct it for these two factors...
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