Three isothermal containers each contain a different perfect...

Three isothermal containers each contain a different perfect gas A, B, and C, and are connected by tubes of negligible volume with closed stopcocks. The pressures and volumes of the three containers are as follows: pA = 1.0 × 10¹ atm, VA = 1.0 × 10¹ L; pB = 4.0 atm, VB = 4.0 L; pC = 1.0 atm, VC = 1.0 L. What will be the final pressure (in atm) if all of the stopcocks connecting the containers are opened? 5.0 5.8 7.8 1.0 10 The mass of the diborane gas in Question 4 is (in g) 2.80 1.72 1568 2.06 0.0204 A sample of diborane gas (B2H6) has a pressure of 345 Torr at -15.0 °C and a volume of 3.48 L. If conditions are changed so that the temperature is 36.0 °C and the pressure is 468 Torr, what will be the final volume (in L) assuming perfect gas conditions? 3.94 2.14 6.16 5.65 3.07

Transcript

00:01 Are if we see we are having perfect gases in this particular case so perfect gases which are here they are a b and c respectively so we are having this pulp a and it is connected with the stopcoke with this bulb b and this bulb b it is further connected with the stovecook to the bulb c so we are considering for the bulb a firstly we can write this thing pressure in the bulb it is 10 atm volume it is also 10 liter here firstly we will be finding moles in the a so from here moles in the bulb by according to the ideal gas equation we know pv it is equal to nrt so from here moles in the a will be pv divided by the rt so from here it can be written as 10 multiplied with this 10 0 .082 multiplied with this t so it is moles in the a now if we see moles in the b chamber, it can be written as again pv divided by the rt.

01:01 So if we say this chamber b have a pressure 4 and volume for this it is also 4.

01:08 So moles in the b chamber, it can be written as 4 multiplied with this 4 divided by the 0 .0821 multiplied with this t.

01:17 So it is a moles in the chamber b.

01:19 Now similarly, moles in the chamber c can be written as pv divided by the rt.

01:25 So from here, if we see pressure and the volume in the t are 1 multiplied with this 1 divided by the 0 .0 8 to 1 multiplied with this t.

01:36 So this is the moles in the c.

01:38 So now if we see, moles which are here initially, moles initially it is equal to the moles finally, because moles are conserved in this particular case.

01:51 So moles initially can be written as moles in the a plus moles in the b plus moles in the c.

01:56 So that can be, we can say, written as 100 divided by the 0 .0821t plus 16 divided by the 0 .0821t plus 1 divided by the 0 .0821t.

02:13 So this is the modes initially.

02:16 And final moles can be written as p -final, v -final divided by the rt.

02:20 So from here, v -final would be 14 -liter as the stop -cox -hires.

02:26 Open so it would be not 14 it would be 15 meters so from here and final can be written as like this we can say p multiplied with this 15 divided by the 0 .08 21 t so this is the we can say final value of this moles now we can say moles initially it is equal to moles final so hundred divided by the 0 .082 1 t plus 16 divided by the 0 .08 to 1 t plus 1 divided by the 0 .08 to 1 t that will be equal to the p multiplied with this 15 0 .082 1 t so from here this thing cancelled throughout so from here 117 it is equal to 15 p so from here p final it would be 7 .8 atm so this is the case 1 now if we see second part here, here for this particular case, we can consider this thing, that pressure multiplied with this volume, it is equal to nrt2.

03:39 So from here, most we can find for the dipurane, pv2, p2 divided by the rt2...

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