Calculate the pressure exerted by 1 mol of an ideal gas in a box that is 0.500 L and 298 K . Hav

step 1 Hi guys, I'm here to solve question 152. This question wants us to calculate the pressure of an

Calculate the pressure exerted by 1 mol of an ideal gas in a...

Key Concepts

Intermolecular Forces

Intermolecular forces are attractive or repulsive forces that act between molecules. These forces play a significant role in determining the behavior of real gases, affecting properties like pressure and volume. In gases with stronger intermolecular attractions, the measured pressure may be lower than predicted by the ideal gas law, while in gases where repulsive forces or molecular volume effects dominate, the pressure might be higher.

Van der Waals Equation

The van der Waals equation modifies the ideal gas law to account for the real behavior of gases by introducing correction terms for the volume occupied by gas molecules and the attractive forces between them. The equation, (P + a(n/V)^2)(V - nb) = nRT, includes the parameters 'a' and 'b' that are specific to each gas and help explain why the pressure measured can be higher or lower than that predicted by the ideal gas law.

Molecular Volume

Molecular volume refers to the space that individual gas molecules occupy. In the ideal gas law, it is assumed that the gas particles have no volume, but in reality, this finite size leads to a reduction in the effective volume available for movement. This concept is incorporated into the van der Waals equation by subtracting the term nb from the volume, thus allowing for a more accurate prediction of the gas's behavior under real conditions.

Ideal Gas Law

The ideal gas law (PV = nRT) relates the pressure, volume, temperature, and amount of a gas under the assumption that there are no intermolecular forces and that the gas molecules occupy negligible space. It provides a simplified model that works well under many conditions, especially at high temperatures and low pressures, where molecules interact minimally.

Real Gas Behavior

Real gases exhibit deviations from the ideal gas behavior under conditions such as high pressures and low temperatures. These deviations occur because of the finite size of the molecules and the presence of intermolecular forces, which cause measurable differences in physical properties compared to predictions made by the ideal gas law.

Transcript

00:02 Hi guys, i'm here to solve question 152.

00:09 This question wants us to calculate the pressure of an id gas.

00:22 They calculate the pressure extracted by 1 mole of 89 id gas in a box that is 0 .5 later and 298 carbon.

00:34 So i have n equals to 1 more and i have v equals 0 .5 later rt temperature is 298 kelvin and i'm here to find my p and mind you this is an id gas so to do this i will use id gas equation which is pv equals nrt so i will write p equals nrt divided by v so i will just substitute one more multiply by r is 0 .082 litre atmosphere per kelvin per more and temperature is 298 kelvin i divide everything by 0 .5 later so from here i get my p to be the 48 .8 7 to atmosphere so let me use red color the right this is the volume of an id gas this is a d gas now i have another question have each group member calculate the pressure of one mode of the following gas gases in the same box at the same temperature we have this gas helium we have neogas hydrogen gas methane and come over the outside gas so we want to calculate their pressure and they have the same one mold the same volume and the same temperature so to do this we use funder wall, each member of the group we use funder wall equation.

03:10 In this case, we are not using id gas law because these are red gases, red gases.

03:23 And there we have some kind of defation from id gas.

03:29 They will defeat a little bit from id gas law.

03:40 So i'm using it so and this is the formula for founder wall equation p plus a n squared divided by v square open the bracket then close the bracket and v minus nb equals nr t you see the difference between p we have we also have this a n square of a few square whereas v also have minus nb of a minus nb so this make a correction correction for intermolecular forces correction for intermolecular forces molecular forces so for id gas load the id gas has assumed that the intermolecular force between the molecule of the gases of gas are negligible so and also this fee and this part and be make a correction for particle volume so i did gas assume that the particle are so small and that their volume is negligible compared to the volume of the container.

05:30 But in the rear, in the rega, the gases have a preciple volume.

05:41 So each member of the group will calculate the pressure using this id and using this one the war equation.

05:54 One member of the group will calculate for lithium, another member will calculate for helium.

06:00 Another member will calculate for hydrogen and we have member that will calculate for pressure for maintain and the last member we calculate for carbon dioxide and you know you have one more so you have your each member so take note you have one more and is one you have your r which is 0 .082 litre atmosphere per capita and you have temperature which is 298 the only thing that each member need now is a is a and b each member needs a and b so this is a constant and from the table 5 .5 i'm going to write it for you guys table 5 .5 from table 5 .5 we have this fellow of a in liter square atmosphere per more and i also have p which is a so you need for ilium which a is 0 .342 and b is 0 .02 and b is 0 .270 for neon as a is 0 .211 b is 0 .0171 for hydrogen a is 0 .244 b is 0 .0266 for mity c h4 a is 2 .25 and b is 0 .0428.

08:42 We have for carbon dioxide.

08:48 So a for carbon dioxide is 3 .59 and b equals 0 .04 to cell.

09:08 Each member of the group we arrange this, find out of our equation is p, you know, fondaway equation is b plus a n square x2v square into v minus n b equals n r t you remember that we all want to find the pressure right if member of the group is looking for pressure p you don't need all this part to do this you can do this by writing equal to so we have n -rptu if you divide by v minus nb to remove this part so you will have v minus nb.

10:15 Then now you are left with plus a n squared by n to remove it to up p a long you subtract it minus a n square defyed by so this is a formula you're going to use p equals n r t defy five by v minus n squared n square defy by v square so each member of the group we use this formula to calculate p so the only thing that will be different for each member is the value of b and a now the first member of the group will calculate for p equals to n r t divided by a minus n square ever three square so to calculate for helium p equals one my n is one that's number of more 0 .082 and t is 298 divided by 0 .5 minus 1 that b value is 0 .3 .7 then minus we have 0 .a a 0 .0 .03 4 2.

12:08 10 is 10 is 1.

12:12 55 by fee 0 .5 square and that gives 51 .38 minus 0 .1368 that is 51 .167 atmosphere.

12:41 This is for ilya.

12:46 This is for earlier.

12:49 So i go back to the second one earlier neo neo okay so p p p we use the same formula when this case is going to be 1 5 0 .082 298 divide by 0 .5 minus 1 times p for idiom 0 .0 .71 then minus a minus a p for aurett and 0 .211 square divided by 0 .5 square...

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