Question:

At 300 K,Carbon powder and O2 are included in container of 10dm3.When the system is heated well,all the O2 converts into CO2.Then the system is cooled to the initial temperature. The volume of the Carbon powder is negligible. Find the percentage of the increasement of the density of the system.

My thoughts on the question:

C(s) + O2(g)--> CO2(g)
300K
ini: y x -

After (y-x) (x-x) (+x)
heating
(moles)

But what happens when the system is cooled again?Do all the CO2 cinvets into O2 so that the moles in the system is same as that of the 300K system?

So the increasement of density=(1/v) { [ (y*12)+(x*32)] - [(x*44)+( (y-x)*12 )] }
Here y cancels out but how do we find a numerical value for the percentage ,when x is there?

(1/v)= 1/volume of the container= 1/(10*(10)^-3)=100 m^(-3)

Yes, mols CO2 is same as mols O2.

I see it this way.
Start with 10 dm3 O2 so end with 10 dm3 CO2.
d O2 to start = 32/22.4 = about 1.42 g/dm3 if at STP. At 300 K it will be 1.42 x 273/300 = ?

CO2 at STP will be 44/22.4 = about 1.96 g/dm3 at STP or 1.96 x 273/300 = ?

Then convert densities into percent change.

Thank you!

To find the percentage of the increase in density of the system after cooling, we need to determine the final density of the system and compare it to the initial density.

First, let's find the initial density of the system. The initial number of moles of oxygen (O2) is x, and the number of moles of carbon dioxide (CO2) is also x because all the oxygen will convert into carbon dioxide. Since the volume of the carbon powder is negligible, the total number of moles in the system is x + x = 2x.

The initial density (ρini) can be calculated by dividing the total mass of the system by the volume of the system. The mass of the system is given by the formula: mass = (number of moles) * (molar mass).

The molar mass of oxygen (O2) is 32 g/mol, and the molar mass of carbon dioxide (CO2) is 44 g/mol. So the initial mass of the system is:
mass = (x * 32 g/mol) + (x * 44 g/mol) = 76x g.

The initial density is then:
ρini = (mass) / (volume) = (76x g) / (10 dm^3) = 7.6x g/dm^3.

Now, let's consider what happens when the system is cooled back to the initial temperature. The carbon dioxide (CO2) does not convert back into oxygen (O2). So we can assume that the number of moles of CO2 remains the same, which is x.

The final density (ρfinal) can be calculated in the same way as the initial density:
mass = (number of moles) * (molar mass).
mass = (x * 44 g/mol) = 44x g.

ρfinal = (mass) / (volume) = (44x g) / (10 dm^3) = 4.4x g/dm^3.

The percentage increase in density is given by the formula:
percentage increase = [(ρfinal - ρini) / ρini] * 100.

Substituting the values, we have:
percentage increase = [(4.4x - 7.6x) / 7.6x] * 100
= [-3.2x / 7.6x] * 100
= -42.11%.

Therefore, the percentage increase in density of the system after cooling is approximately -42.11%. The negative sign indicates a decrease in density compared to the initial density.