CBSE Class 11 Physics Chapter 13 Kinetic Theory NCERT Solutions

We are given the diameter of an oxygen molecule, $d\; =\; 3\; \backslash text\{\; \AA \}$. The radius $r$ is half of the diameter:

$r\; =\; \backslash frac\{d\}\{2\}\; =\; \backslash frac\{3\}\{2\}\; =\; 1.5\; \backslash text\{\; \AA \}$

To work with standard units, we convert the radius to centimeters: $1.5\; \backslash text\{\; \AA \}\; =\; 1.5\; \backslash times\; 10^\{-8\}\; \backslash text\{\; cm\}$.

The volume of a single oxygen molecule, assuming it to be spherical, is given by the formula for the volume of a sphere:

$$V\_\{\backslash text\{molecule\}\}\; =\; \backslash frac\{4\}\{3\}\backslash pi\; r^3$$

The total molecular volume occupied by 1 mole of oxygen gas is the volume of a single molecule multiplied by Avogadro's number ($N\_A\; =\; 6.023\; \backslash times\; 10^\{23\}\; \backslash text\{\; molecules/mole\}$):

$$V\_\{\backslash text\{molecular\}\}\; =\; N\_A\; \backslash times\; \backslash frac\{4\}\{3\}\backslash pi\; r^3$$

Substituting the values:

$$V\_\{\backslash text\{molecular\}\}\; =\; (6.023\; \backslash times\; 10^\{23\})\; \backslash times\; \backslash frac\{4\}\{3\}\; \backslash times\; 3.14\; \backslash times\; (1.5\; \backslash times\; 10^\{-8\}\; \backslash text\{\; cm\})^3$$

$$V\_\{\backslash text\{molecular\}\}\; =\; (6.023\; \backslash times\; 10^\{23\})\; \backslash times\; \backslash frac\{4\}\{3\}\; \backslash times\; 3.14\; \backslash times\; (3.375\; \backslash times\; 10^\{-24\}\; \backslash text\{\; cm\}^3)$$

$$V\_\{\backslash text\{molecular\}\}\; \backslash approx\; 8.51\; \backslash text\{\; cm\}^3$$

At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22400 cm³ (or 22.4 litres). This is the actual volume occupied by the gas.

The fraction of molecular volume to the actual volume is:

$$\backslash text\{Fraction\}\; =\; \backslash frac\{V\_\{\backslash text\{molecular\}\}\}\{V\_\{\backslash text\{actual\}\}\}$$

$$\backslash text\{Fraction\}\; =\; \backslash frac\{8.51\; \backslash text\{\; cm\}^3\}\{22400\; \backslash text\{\; cm\}^3\}$$

$$\backslash text\{Fraction\}\; \backslash approx\; 3.8\; \backslash times\; 10^\{-4\}$$

Thus, the fraction of molecular volume to the actual volume occupied by oxygen gas at STP is approximately $3.8\; \backslash times\; 10^\{-4\}$.

We can use the ideal gas equation to determine the molar volume at STP. The ideal gas equation is given by:

$$PV\; =\; nRT$$

Where:

• $P$ is the pressure of the gas.
• $V$ is the volume of the gas.
• $n$ is the number of moles of the gas.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature of the gas.

At Standard Temperature and Pressure (STP), the conditions are:

• Standard Temperature ($T$) = 0 °C = 273.15 K (we use 273 K for approximation as in the source).
• Standard Pressure ($P$) = 1 atm = $1.013\; \backslash times\; 10^5\; \backslash text\{\; N\; m\}^\{-2\}$.
• Number of moles ($n$) = 1 mol (for molar volume).

The universal gas constant ($R$) has a value of $8.314\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\}$.

We need to find the volume ($V$). Rearranging the ideal gas equation to solve for $V$:

$$V\; =\; \backslash frac\{nRT\}\{P\}$$

Now, substitute the values for STP conditions:

$$V\; =\; \backslash frac\{(1\; \backslash text\{\; mol\})\; \backslash times\; (8.314\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\})\; \backslash times\; (273\; \backslash text\{\; K\})\}\{1.013\; \backslash times\; 10^5\; \backslash text\{\; N\; m\}^\{-2\}\}$$

Calculating the value:

$$V\; \backslash approx\; \backslash frac\{2270.082\}\{1.013\; \backslash times\; 10^5\}\; \backslash text\{\; m\}^3$$

$$V\; \backslash approx\; 0.0224\; \backslash text\{\; m\}^3$$

Since 1 cubic meter ($m^3$) is equal to 1000 litres, we convert the volume to litres:

$$V\; \backslash approx\; 0.0224\; \backslash times\; 1000\; \backslash text\{\; litres\}$$

$$V\; \backslash approx\; 22.4\; \backslash text\{\; litres\}$$

Therefore, it is shown that the molar volume of an ideal gas at STP is approximately 22.4 litres.

The question refers to a figure (Figure 13.8) which is not provided in the source text. However, based on the description, the plot shows the relationship between $\backslash frac\{PV\}\{T\}$ and $P$ for a fixed mass of oxygen gas at two different temperatures.

From the ideal gas law, $PV\; =\; nRT$. We can rewrite this as $\backslash frac\{PV\}\{T\}\; =\; nR$.

For a fixed mass of gas (like $1.00\; \backslash times\; 10^\{-3\}$ kg of oxygen), the number of moles ($n$) is constant. The universal gas constant ($R$) is also a constant.

Therefore, the product $nR$ is a constant value for a given amount of gas. This means that $\backslash frac\{PV\}\{T\}$ should be constant, regardless of the pressure ($P$), as long as the temperature is constant.

The plot described would ideally show a horizontal line (constant value of $\backslash frac\{PV\}\{T\}$) versus pressure ($P$). If the plot shows two different temperatures, it implies that the value of $\backslash frac\{PV\}\{T\}$ might be different for different temperatures if the mass of gas changes, or if the gas deviates from ideal behavior. However, for a fixed mass of gas, $n$ is constant, so $nR$ is constant. The plot of $\backslash frac\{PV\}\{T\}$ vs $P$ for a fixed mass of gas should be a horizontal line, indicating that $\backslash frac\{PV\}\{T\}$ is independent of pressure.

The figure likely shows two such horizontal lines, one for each temperature, possibly at different heights if the number of moles $n$ were different, but since the mass is fixed, $n$ is fixed, and thus $nR$ is fixed. The plot should ideally be a single horizontal line if the gas behaves ideally.

Without the figure, a detailed interpretation or calculation is not possible. However, the principle is that $\backslash frac\{PV\}\{T\}$ is proportional to the number of moles ($n$) for an ideal gas.