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 the diameter:

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

To calculate the volume of a single molecule, we first convert the radius to centimeters:

$$r\; =\; 1.5\; \backslash times\; 10^\{-8\}\; \backslash text\{\; cm\}$$

The volume of a single oxygen molecule (approximated as a sphere) is given by the formula for the volume of a sphere, $V\_\{molecule\}\; =\; \backslash frac\{4\}\{3\}\backslash pi\; r^3$. However, the question asks for the total molecular volume occupied by 1 mole of oxygen gas. This is calculated by multiplying the volume of a single molecule by Avogadro's number ($N\_A$).

Avogadro's number, $N\_A\; =\; 6.023\; \backslash times\; 10^\{23\}$ molecules/mole.

The total molecular volume ($V\_\{molecular\}$) for 1 mole of oxygen is:

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

Substituting the values:

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

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

The actual volume occupied by 1 mole of any ideal gas at Standard Temperature and Pressure (STP) is approximately 22400 $cm^3$.

The fraction of molecular volume to the actual volume is:

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

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

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

Therefore, 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
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the universal gas constant
• $T$ is the absolute temperature

For standard temperature and pressure (STP):

• Number of moles, $n\; =\; 1\; \backslash text\{\; mol\}$
• Standard temperature, $T\; =\; 0^\backslash circ\; \backslash text\{C\}\; =\; 273.15\; \backslash text\{\; K\}$ (often approximated as 273 K for calculations)
• Standard pressure, $P\; =\; 1\; \backslash text\{\; atm\}$

The universal gas constant $R$ has a value of $8.314\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\}$. We need to express pressure in Pascals (Pa) for consistency with SI units of R. $1\; \backslash text\{\; atm\}\; =\; 1.013\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 1.013\; \backslash times\; 10^5\; \backslash text\{\; N\; m\}^\{-2\}$.

Now, we rearrange the ideal gas equation to solve for volume ($V$):

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

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:

$$V\; \backslash approx\; \backslash frac\{2270.562\}\{1.013\; \backslash times\; 10^5\}\; \backslash text\{\; m\}^3$$

$$V\; \backslash approx\; 0.02241\; \backslash text\{\; m\}^3$$

To convert this volume to litres, we use the conversion factor $1\; \backslash text\{\; m\}^3\; =\; 1000\; \backslash text\{\; litres\}$:

$$V\; \backslash approx\; 0.02241\; \backslash times\; 1000\; \backslash text\{\; litres\}$$

$$V\; \backslash approx\; 22.41\; \backslash text\{\; litres\}$$

Thus, 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 text. However, we can discuss the expected behavior of the plot based on the ideal gas law. The ideal gas law is given by $PV\; =\; nRT$. We can rearrange this equation to express $PV/T$ as a function of pressure $P$ and temperature $T$.

From $PV\; =\; nRT$, we get:

$$\backslash frac\{PV\}\{T\}\; =\; nR$$

Here, $n$ is the number of moles of the gas, and $R$ is the universal gas constant. For a fixed amount of gas (i.e., constant $n$), the product $nR$ is a constant.

The mass of oxygen gas given is $1.00\; \backslash times\; 10^\{-3\}\; \backslash text\{\; kg\}$. The molar mass of oxygen ($O\_2$) is approximately $32\; \backslash text\{\; g/mol\}\; =\; 0.032\; \backslash text\{\; kg/mol\}$.

The number of moles $n$ can be calculated as:

$$n\; =\; \backslash frac\{\backslash text\{mass\}\}\{\backslash text\{molar\; mass\}\}\; =\; \backslash frac\{1.00\; \backslash times\; 10^\{-3\}\; \backslash text\{\; kg\}\}\{0.032\; \backslash text\{\; kg/mol\}\}\; \backslash approx\; 0.03125\; \backslash text\{\; mol\}$$

Therefore, for this specific amount of oxygen gas, $nR$ is a constant value:

$$nR\; =\; (0.03125\; \backslash text\{\; mol\})\; \backslash times\; (8.314\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\})\; \backslash approx\; 0.26\; \backslash text\{\; J\; K\}^\{-1\}$$

So, the equation for the plot is:

$$\backslash frac\{PV\}\{T\}\; =\; \backslash text\{constant\}\; \backslash quad\; (\backslash text\{for\; a\; fixed\; mass\; of\; gas\})$$

This implies that for a fixed mass of gas, the quantity $PV/T$ should remain constant, regardless of the pressure $P$ or temperature $T$, as long as the gas behaves ideally. Thus, the plot of $PV/T$ versus $P$ should be a horizontal line.

The question states that the plot is shown at two different temperatures. If the gas behaves ideally, the plot of $PV/T$ versus $P$ should be a horizontal line, meaning $PV/T$ is independent of $P$ and $T$ for a fixed amount of gas. However, real gases deviate from ideal behavior, especially at high pressures and low temperatures. The figure likely shows slight deviations or is intended to illustrate the ideal gas law principle.

If the figure shows two different horizontal lines at different heights, it would contradict the ideal gas law for a fixed amount of gas, as $nR$ should be constant. If the figure shows two horizontal lines at the same height, it confirms the ideal gas behavior where $PV/T$ is constant for a fixed mass.

Without the figure, a precise interpretation is not possible. However, based on the ideal gas law, the plot of $PV/T$ versus $P$ for a fixed mass of gas should be a horizontal line, indicating that $PV/T$ is constant.