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**Hint:**We know that the ideal gas law is the condition of a speculative ideal gas. It is a decent estimation of the conduct of numerous gases under numerous conditions, despite the fact that it has a few impediments

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.

**Complete step by step answer:**

Given,

The number of moles $n = 10$.

The pressure of gas is $0.821atm$.

The value of gas constant is $0.0821Latmmo{l^{ - 1}}{K^{ - 1}}$.

We know the ideal gas equation is,

$PV = nRT$

Now, substitute the known qualities in the above condition,

$V = \dfrac{{10 \times 0.0821 \times T}}{{0.821}}$

$V = 1 \times T$

$\dfrac{{\log V}}{{\log T}} = 1$

$\tan \theta = 1$

$\tan {45^ \circ } = 1$

**Hence option A is correct.**

**Additional information:**

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.

**Note:**

We know that,

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

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

${\text{Density = }}\dfrac{{\text{n}}}{{{\text{volume}}}}$

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,

\[{\text{Density = }}\dfrac{{P\not V}}{{RT\not V}}\]

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

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

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

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