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The ideal gas equation is,

$PV = nRT$

Where P is the pressure in the atmosphere.

V is the volume of gas in a liter.

n is the number of moles.

R is a universal gas constant.

T is the temperature.

We know that $Kp = PC{O_2}$

It is given that value of $Kp = 1.16atm$

The temperature is ${800^ \circ }C$ .

The volume is $10L$ .

The gas constant is $R = 0.082Latm/mol/K$

Now we can calculate the number of moles reacted using the ideal gas equation,

\[{\text{n = }}\dfrac{{PV}}{{RT}}\]

\[\dfrac{{\text{W}}}{M}{\text{ = }}\dfrac{{1.16 \times 10}}{{0.082 \times 1073}}\]

\[{\text{W = }}\dfrac{{1.16 \times 10 \times 44}}{{0.082 \times 1073}}\]

On simplifying we get,

\[{\text{W = 5}}{\text{.8g}}\]

The balanced equation is,

$CaC{O_3}\left( s \right) \rightleftharpoons CaO\left( s \right) + C{O_2}$

The mass of calcium carbonate reacted can be calculated as,

$\dfrac{x}{{100}} = \dfrac{{5.8g}}{{44}}$

On simplifying we get,

$x = 13.1g$

The mass of calcium carbonate unreacted $ = 20 - 13.1 = 6.9$

The percentage of calcium carbonate unreacted $ = \dfrac{{6.9}}{{20}} \times 100 = 34.5\% $

If the gas obeys an ideal gas equation then the pressure is given by,

${\text{P = }}\dfrac{{{\text{nRT}}}}{{\text{V}}} \to (1)$

If the volume is doubled and the temperature is halved then the equation becomes,

${\text{P = }}\dfrac{{{\text{nRT/2}}}}{{{\text{2V}}}}$

${\text{P = }}\dfrac{{{\text{nRT}}}}{{{\text{4V}}}} \to 2$

From equation 1 ${\text{P = }}\dfrac{{{\text{nRT}}}}{{\text{V}}}$ then the equation 2 becomes,

${\text{P = }}\dfrac{{\text{P}}}{{\text{4}}}$

Thus, if the volume is doubled and the temperature is halved then the pressure of the system decreases by four times.

We know that,

$Density = \dfrac{{mass}}{{volume}}$

Assuming mass is equal to the number of moles in ideal gas.

$Density = \dfrac{n}{{Volume}}$

The ideal gas equation is,

${\text{PV = nRT}}$

The number of moles can be calculated as,

${\text{n = }}\dfrac{{{\text{PV}}}}{{{\text{RT}}}}$

Substituting the value of n in density equation,

\[{\text{Density = }}\dfrac{{PV}}{{RTV}}\]

\[{\text{Density = }}\dfrac{{\text{P}}}{{{\text{RT}}}}\]

\[{\text{Density}} \propto \dfrac{{\text{1}}}{{\text{T}}}\]

It is clear that density is inversely proportional to temperature. Thus, as the density of the gas decreases temperature increases.